Student Solutions - Complete K-12 Math Curriculum · Student Solutions stUDent soLUtions ... answer...
108
LESSON PRACTICE 1A - SYSTEMATIC REVIEW 1D GEOMETRY 133 Lesson Practice 1A 1. 2. starting point or origin of a ray contained in the same line having length, but no width 3. 4. 5. 6. an infinite number of connected points ray line segment equal similar congruent fals 7. 8. 9. 10. e: The common endpoint is B. true: li 11. 12. true ne BC extends indefinitely in both directions, so it includes They have only one AB false . : 13. point in common. true: ray BC extends indef 14. initely to the right, so it includes everything in that direction. two quantities 15. 16. = ( ) ≅ two geometric figures two quantities ( ) = ( ) 17. 18. 19. ≅ ( ) ≅ two geometric figures two geometric figures two quantities ( ) = ( ) 20. Lesson Practice 1B 1. 2. 3. 4. 5. 6. geometry point line collinear ray segment equal congruent 7. 8. 9. similar endpoint 10. 11. line 12. 13. 14. 15. 16. ray line segment congruent point C ray DE lines AB or BC or AC or BA or CB or C 17. A an infinite number rays BA 18. 19. 20. A or B or C or CA or CB Systematic Review 1C 1. 2. 3. 4. 5. ray segment , shape, size point line points ray line segment line, congruent to, line 6. 7. 8. 9. 10. 11. ray geometry has same shape but different size exactly the same length or measure 12. 13. in the same line exactly the same shape and 14. size point S rays MP or MQ RS Any 15. 16. 17. line : answer that refers to this line is acceptable. 18. 19. 20. M P orQ , , infinite infinite Systematic Review 1D 1. 2. line both are infi same : nite line segment congruent, equal 3. 4. Student Solutions STUDENT SOLUTIONS Solutions are shown in detail. The student may use canceling and other shortcuts as long as the answers match. If you see an error, see online solutions mentioned on page 14.
Transcript of Student Solutions - Complete K-12 Math Curriculum · Student Solutions stUDent soLUtions ... answer...
Lesson Practice 1a - sYsteMatic reVieW 1DGeoMetrY 133
Lesson Practice 1A1.
2.
starting point or origin of a ray
contained iin the same line
having length, but no width3.
4..
5.
6.
an infinite number
of connected points
ray
lline segment
equal
similar
congruent
fals
7.
8.
9.
10. ee: the common endpoint is B.
true: li
11.
12.
true
nne Bc extends indefinitely in
both directions, sso it includes
they have only one
aB
false
.
:13. point
in common.
true: ray Bc extends indef14. iinitely to
the right, so it includes
everythingg in that direction.
two quantities15.
16.
= ( )≅ ttwo geometric figures
two quantities
( )= ( )17.
18..
19.
≅ ( )≅
two geometric figures
two geometric fiigures
two quantities
( )= ( )20.
Lesson Practice 1B1.
2.
3.
4.
5.
6.
geometry
point
line
collinear
ray
segmentt
equal
congruent
endpoint
7.
8.
9.
10.
11.
similar
line
112.
13.
14.
15.
16.
ray
line segment
congruent
point c
rray De
lines aB or Bc or ac or
Ba or cB or c
17.
aa
an infinite number
rays Ba
18.
19.
20.
a or B or c
or ca or cB
1.
2.
3.
4.
5.
6.
geometry
point
line
collinear
ray
segmentt
equal
congruent
endpoint
7.
8.
9.
10.
11.
similar
line
112.
13.
14.
15.
16.
ray
line segment
congruent
point c
rray De
lines aB or Bc or ac or
Ba or cB or c
17.
aa
an infinite number
rays Ba
18.
19.
20.
a or B or c
or ca or cB
Systematic Review 1C 1.
2.
3.
4.
5.
ray segment,
shape, size
point
line
pointss
ray
line segment
line, congruent to, line
6.
7.
8.
99.
10.
11.
ray
geometry
has same shape but differentt size
exactly the same length or measure12.
13. iin the same line
exactly the same shape and 14. ssize
point s
rays MP or MQ
rs any
15.
16.
17. line : aanswer that refers
to this line is acceptable.
118.
19.
20.
M P or Q, ,
infinite
infinite
Systematic Review 1DSystematic Review 1D1.
2.
line
both are infisame : nnite
line segment
congruent, equal
a
measu
3.
4.
5.
6. rre, earth
point
similar
collinear
points
7.
8.
9.
10.
111.
12.
line aB is congruent to line cD
distance aBB is equal to distance cD
line segment aB is13. congruent to
line segment cD
ray aB is congr14. uuent to ray cD
alse: they do not lie on
the
15. f
ssame line.
alse: they have only one
16.
17.
true
f ppoint
in common.
alse: they have no
common e
18. f
nndpoint.
rue: they both refer to the
same li
19. t
nne segment.
rue: the line is not drawn,
but
20. t
iit could be.
Student Solutions
stUDent soLUtions
Solutions are shown in detail. The student may use canceling and other shortcuts as long as the answers match. If you see an error, see online solutions mentioned on page 14.
sYsteMatic reVieW 1D - sYsteMatic reVieW 2c
soLUtions GeoMetrY134
Systematic Review 1D1.
2.
line
both are infisame : nnite
line segment
congruent, equal
a
measu
3.
4.
5.
6. rre, earth
point
similar
collinear
points
7.
8.
9.
10.
111.
12.
line aB is congruent to line cD
distance aBB is equal to distance cD
line segment aB is13. congruent to
line segment cD
ray aB is congr14. uuent to ray cD
alse: they do not lie on
the
15. f
ssame line.
alse: they have only one
16.
17.
true
f ppoint
in common.
alse: they have no
common e
18. f
nndpoint.
rue: they both refer to the
same li
19. t
nne segment.
rue: the line is not drawn,
but
20. t
iit could be.
Lesson Practice 2ALesson Practice 2A1.
2.
3.
length and width
two
same
44.
5.
6.
two
meet
co
–dimensional; three–dimensional
mmbined
collection
null
plane
sub
7.
8.
9.
10.
or group
sset
union
−
− ∅
− ∪
− ∩
11.
12.
13.
null set
intersection
114.
15.
16.
true
false
false t
ray Be ray BF point B∩ =
: hhe line segments have no
intersection, but theirr union is
simply the two segments.
17.
18.
true
fallse : of the points mentioned,
only B is a subsett of line eF.
Lesson Practice 2A1.
2.
3.
length and width
two
same
44.
5.
6.
two
meet
co
–dimensional; three–dimensional
mmbined
collection
null
plane
sub
7.
8.
9.
10.
or group
sset
union
−
− ∅
− ∪
− ∩
11.
12.
13.
null set
intersection
114.
15.
16.
true
false
false t
ray Be ray BF point B∩ =
: hhe line segments have no
intersection, but theirr union is
simply the two segments.
17.
18.
true
fallse : of the points mentioned,
only B is a subsett of line eF.
Lesson Practice 2BLesson Practice 2B1.
2.
3.
4.
point
line
plane
coplanarr
set
plane
empty or null set
three
a is
5.
6.
7.
8.
9. aa subset of B
the union of a and B
the int
10.
11. eersection of a and B
the set containing a an12. dd B
a is an empty set.
false: the union is
13.
14. the
two segments.
false: only s is containe15. dd
in the intersection.
false:
16.
17.
18.
true
true
QQ is not contained in rt.
Systematic Review 2CSystematic Review 2C1.
2.
3.
plane
coplanar
collinearr
similar
intersection
union
congruent
se
4.
5.
6.
7.
8. tt
empty or null set
equal
union
null or
9.
10.
11.
12. empty set
subset
intersection
answe
13.
14.
15. any rr representing ray DF
any lines or rays thator contain De
be correct.
answer repr
would
any16. eesenting line aB
would be correct.
or 17.
18
Bc cB
..
19.
20.
Be eB
any
or
empty or null set
answer reprresenting line FD
would be correct.
sYsteMatic reVieW 2c - Lesson Practice 3a
soLUtionsGeoMetrY 135
Systematic Review 2C1.
2.
3.
plane
coplanar
collinearr
similar
intersection
union
congruent
se
4.
5.
6.
7.
8. tt
empty or null set
equal
union
null or
9.
10.
11.
12. empty set
subset
intersection
answe
13.
14.
15. any rr representing ray DF
any lines or rays thator contain De
be correct.
answer repr
would
any16. eesenting line aB
would be correct.
or 17.
18
Bc cB
..
19.
20.
Be eB
any
or
empty or null set
answer reprresenting line FD
would be correct.
Systematic Review 2DSystematic Review 2D1.
2.
3
empty or null set
plane
..
4.
5.
6.
endpoint, origin
intersection
subset
union
77.
8.
9.
10.
line
point
congruent
two lines in the samme plane
two points on the same line
two s
11.
12. qquares with
different dimensions
two squares13. with same dimensions
two measurements
with
14.
tthe same value
L
one
e
15.
16.
17.
18.
19.
eH
eL or eH→→
nndpoint, origin
K, L, G20.
Systematic Review 2D1.
2.
3
empty or null set
plane
..
4.
5.
6.
endpoint, origin
intersection
subset
union
77.
8.
9.
10.
line
point
congruent
two lines in the samme plane
two points on the same line
two s
11.
12. qquares with
different dimensions
two squares13. with same dimensions
two measurements
with
14.
tthe same value
L
one
e
15.
16.
17.
18.
19.
eH
eL or eH→→
nndpoint, origin
K, L, G20.
Systematic Review 2ESystematic Review 2E1.
2.
3.
4.
set
or
ray
∅ { },
line ssegment
union
intersection
collinear
e
5.
6.
7.
8.
9.
1
00. infinite although only 5 points are
labeled
;
,, any plane contains an
infinite number of poinnts.
es. two lines that intersect are
in th
11. y
ee same plane.
infinite; although only 3 poin12. tts are
labeled, any line contains ann infinite
number of points.
infi13. nnite
es. any two points can be
connected by
14. y
a straight line.
it does not lie i
15.
16.
no
no. nn plane x given
ray ae
( )
×( ) × − =(
.
17.
18. 2 5 42 52
10)) × − =
− =
× ( ) =
× ( )
16 25
160 25 135
42 3 62 12
42 3 36 12
19. ÷ ÷
÷ ÷ ==
× ==
+ =+ =+ =
42 3 3126 3 42
28 22 62
28 4 367 36 43
÷÷
÷÷
20.
Systematic Review 2E1.
2.
3.
4.
set
or
ray
∅ { },
line ssegment
union
intersection
collinear
e
5.
6.
7.
8.
9.
1
00. infinite although only 5 points are
labeled
;
,, any plane contains an
infinite number of poinnts.
es. two lines that intersect are
in th
11. y
ee same plane.
infinite; although only 3 poin12. tts are
labeled, any line contains ann infinite
number of points.
infi13. nnite
es. any two points can be
connected by
14. y
a straight line.
it does not lie i
15.
16.
no
no. nn plane x given
ray ae
( )
×( ) × − =(
.
17.
18. 2 5 42 52
10)) × − =
− =
× ( ) =
× ( )
16 25
160 25 135
42 3 62 12
42 3 36 12
19. ÷ ÷
÷ ÷ ==
× ==
+ =+ =+ =
42 3 3126 3 42
28 22 62
28 4 367 36 43
÷÷
÷÷
20.
Lesson Practice 3ALesson Practice 3A1.
2.
3.
4.
5.
6.
t
t
B
t
rtQ or Qtr
W
∠ ∠
77.
8.
9.
∠ ∠XWV or VWX
W
º116 You may want to accept
answers that are a degree or two
either way foor this and similar
problems.
10.
11.
12.
24
90
75
º
º
ºº
º
−
−
check with protractor
check with p13. 95 rrotractor
check with protractor14. 170º −
Lesson Practice 3a - sYsteMatic reVieW 3e
soLUtions GeoMetrY136
Lesson Practice 3A1.
2.
3.
4.
5.
6.
t
t
B
t
rtQ or Qtr
W
∠ ∠
77.
8.
9.
∠ ∠XWV or VWX
W
º116 You may want to accept
answers that are a degree or two
either way foor this and similar
problems.
10.
11.
12.
24
90
75
º
º
ºº
º
−
−
check with protractor
check with p13. 95 rrotractor
check with protractor14. 170º −
Lesson Practice 3BLesson Practice 3B1.
2.
3.
4.
5.
6.
J
G
s
H
cHB or BHc
M
∠ ∠
77.
8.
9.
10.
11.
12.
∠ ∠
−
XMY or YMX
M
º
º
º
º
138
49
24
15 cheeck with protractor
check with protra13. 160º − cctor
check with protractor14. 110º −
Lesson Practice 3B1.
2.
3.
4.
5.
6.
J
G
s
H
cHB or BHc
M
∠ ∠
77.
8.
9.
10.
11.
12.
∠ ∠
−
XMY or YMX
M
º
º
º
º
138
49
24
15 cheeck with protractor
check with protra13. 160º − cctor
check with protractor14. 110º −
Systematic Review 3CSystematic Review 3C1.
2.
3.
4.
angles
vertex
M
measuree, angle
false: the inter
5.
6.
7.
∠ ∠BMc or cMB
true
ssection of
two planes is a line
false: the int8. eersection of
two lines is a point
9.
10.
11
true
73º
..
12.
13.
14.
101
12
30
25
º
º
º − check with protractor
ºº
º
−
−
check with protractor
check with p15. 90 rrotractor
rays Da
16.
17.
18.
19.
20.
B
ray cD
or Dc
c
∅
Systematic Review 3C1.
2.
3.
4.
angles
vertex
M
measuree, angle
false: the inter
5.
6.
7.
∠ ∠BMc or cMB
true
ssection of
two planes is a line
false: the int8. eersection of
two lines is a point
9.
10.
11
true
73º
..
12.
13.
14.
101
12
30
25
º
º
º − check with protractor
ºº
º
−
−
check with protractor
check with p15. 90 rrotractor
rays Da
16.
17.
18.
19.
20.
B
ray cD
or Dc
c
∅
Systematic Review 3DSystematic Review 3D1.
2.
3.
degrees
vertex
coplanar
44.
5.
6.
7.
8.
measure, angle alpha
similar
90
51
170
º
º
ºº
º
º
9.
10.
179
18
−
−
check with protractor
check wwith protractor
check with protractor11.
1
88º −
22.
13.
infinite
every plane is two-dimensiona2 − ll
line cD
line eG or any other answer
14.
15. that
refers to the same line
infini
16.
17.
18.
c
x
tte: every plane contains an
infinite number of ppoints
19.
20.
cD
line cD
Systematic Review 3D1.
2.
3.
degrees
vertex
coplanar
44.
5.
6.
7.
8.
measure, angle alpha
similar
90
51
170
º
º
ºº
º
º
9.
10.
179
18
−
−
check with protractor
check wwith protractor
check with protractor11.
1
88º −
22.
13.
infinite
every plane is two-dimensiona2 − ll
line cD
line eG or any other answer
14.
15. that
refers to the same line
infini
16.
17.
18.
c
x
tte: every plane contains an
infinite number of ppoints
19.
20.
cD
line cD
Systematic Review 3ESystematic Review 3E1.
2.
3.
true
true
false: the uniion is eF.
true
true: While this line
4.
5.
6.
false
is not shown,
such a line could be drawn.
7. truee
8.
9.
true
false: they have no common
end point.
110.
11.
12.
13.
16
90
122
13
º
º
º
º − check with protracttor
check with protractor
ch
14.
15.
125
170
º
º
−
− eeck with protractor
true: commutative proper16. tty
of addition
true: commutative property
of
17.
multiplication
false: Division is not commu18. ttative.
false: subtraction is
not commutativ
19.
ee.
true: commutative property
of addition
20.
sYsteMatic reVieW 3e - sYsteMatic reVieW 4D
soLUtionsGeoMetrY 137
Systematic Review 3E1.
2.
3.
true
true
false: the uniion is eF.
true
true: While this line
4.
5.
6.
false
is not shown,
such a line could be drawn.
7. truee
8.
9.
true
false: they have no common
end point.
110.
11.
12.
13.
16
90
122
13
º
º
º
º − check with protracttor
check with protractor
ch
14.
15.
125
170
º
º
−
− eeck with protractor
true: commutative proper16. tty
of addition
true: commutative property
of
17.
multiplication
false: Division is not commu18. ttative.
false: subtraction is
not commutativ
19.
ee.
true: commutative property
of addition
20.
Lesson Practice 4ALesson Practice 4A1.
2.
3.
4.
acute
obtuse
right
acute
55.
6.
7.
8.
acute
180º, straight
270º, reflex
90º, rigght
reflex
acute
reflex
obtuse
9.
10.
11.
12.
Lesson Practice 4A1.
2.
3.
4.
acute
obtuse
right
acute
55.
6.
7.
8.
acute
180º, straight
270º, reflex
90º, rigght
reflex
acute
reflex
obtuse
9.
10.
11.
12.
Lesson Practice 4BLesson Practice 4B1.
2.
3.
4.
straight
obtuse
reflex
riight
reflex
56º, acute
35º, acute
136º, ob
5.
6.
7.
8. ttuse
obtuse
acute
right
obtuse
9.
10.
11.
12.
Lesson Practice 4B1.
2.
3.
4.
straight
obtuse
reflex
riight
reflex
56º, acute
35º, acute
136º, ob
5.
6.
7.
8. ttuse
obtuse
acute
right
obtuse
9.
10.
11.
12.
Systematic Review 4CSystematic Review 4C1.
2.
∠ ∠
∠ ∠ ∠
aeB or Bea
, , ,α β γ , , ,
, , ,
∠ ∠ ∠ ∠
∠ ∠ ∠ ∠
aeB Bec ace
ceD aBe Bce aec
BeD,
º
, ,
or ecD
aec
Bae eDc aeD or acD
∠
∠
∠ ∠ ∠ ∠
3.
4.
5.
6
90
..
7.
8.
9.
10.
11.
12
e
BeD
eBc
acute
or DeB
acute
obtuse
∠
..
13.
14.
15.
collinear
earth, measure
congruent
: n∅ uull or empty set
false: a point has z
16.
17.
true
eero dimensions.
false: a plane has two dimen18. ssions.
true: any angle between
90º and 180º
19.
iis obtuse.
false: a line segment
has definit
20.
ee length.
Systematic Review 4C1.
2.
∠ ∠
∠ ∠ ∠
aeB or Bea
, , ,α β γ , , ,
, , ,
∠ ∠ ∠ ∠
∠ ∠ ∠ ∠
aeB Bec ace
ceD aBe Bce aec
BeD,
º
, ,
or ecD
aec
Bae eDc aeD or acD
∠
∠
∠ ∠ ∠ ∠
3.
4.
5.
6
90
..
7.
8.
9.
10.
11.
12
e
BeD
eBc
acute
or DeB
acute
obtuse
∠
..
13.
14.
15.
collinear
earth, measure
congruent
: n∅ uull or empty set
false: a point has z
16.
17.
true
eero dimensions.
false: a plane has two dimen18. ssions.
true: any angle between
90º and 180º
19.
iis obtuse.
false: a line segment
has definit
20.
ee length.
Systematic Review 4DSystematic Review 4D1.
2.
3.
4.
acute
obtuse
reflex
rigght
180º
Ptr or rtP
lines Ms,
5.
6.
7.
8.
9.
10.
90
90
º
º
t
st or Mt
right
acute
obtuse
straight
11.
12.
13.
14.
155.
16.
acute
see drawing
labeling of lines can be sswitched
infinite
see drawing
( )
∠ ∠
17.
18.
19. ceB B, eeD Dea aec
e
, ,∠ ∠
20.
sYsteMatic reVieW 4D - Lesson Practice 5a
soLUtions GeoMetrY138
Systematic Review 4D1.
2.
3.
4.
acute
obtuse
reflex
rigght
180º
Ptr or rtP
lines Ms,
5.
6.
7.
8.
9.
10.
90
90
º
º
t
st or Mt
right
acute
obtuse
straight
11.
12.
13.
14.
155.
16.
acute
see drawing
labeling of lines can be sswitched
infinite
see drawing
( )
∠ ∠
17.
18.
19. ceB B, eeD Dea aec
e
, ,∠ ∠
20.
A
B
CD
Ex
Systematic Review 4ESystematic Review 4E1.
2.
3.
4.
right
obtuse
acute
straaight
reflex
Z
H
5.
6.
7.
8.
9.
10.
H
BHe or eHB
YPs or
∠ ∠
∠ ∠sPY
11.
12.
13.
14.
15.
obtuse
obtuse
straight
reflex
ffalse: a line is one-dimensional.
true
fal
16.
17. sse: a point has neither
length nor width.
tru18. ee
19.
20.
30 42 1830 18 42
12 42124227
15 4
= +− =
=
=
=
= −
YYY
Y
Y
55 3015 30 45
45 454545
1
MMM
M
M
−+ = −
= −
−=
− =
Systematic Review 4E1.
2.
3.
4.
right
obtuse
acute
straaight
reflex
Z
H
5.
6.
7.
8.
9.
10.
H
BHe or eHB
YPs or
∠ ∠
∠ ∠sPY
11.
12.
13.
14.
15.
obtuse
obtuse
straight
reflex
ffalse: a line is one-dimensional.
true
fal
16.
17. sse: a point has neither
length nor width.
tru18. ee
19.
20.
30 42 1830 18 42
12 42124227
15 4
= +− =
=
=
=
= −
YYY
Y
Y
55 3015 30 45
45 454545
1
MMM
M
M
−+ = −
= −
−=
− =
Lesson Practice 5ALesson Practice 5A1.
2.
3.
parallel
perpendicular
bissector
perpendicular bisector
midpoint
Foll
4.
5.
6. oow the procedure in the text.
Use a ruler to chheck that the line
segments of each side of thee
bisector have equal lengths.
Follow the pro7. ccedure in the text.
Use a ruler to check that tthe line
segments of each side of the
bisector have equal lengths.
Follow the procedure in 8. tthe text.
Use a protractor to check that the anngles
on each side of the bisector have
equal meeasures.
Follow the procedure in the text.
Us
9.
ee a protractor to check that the angles
on each side of the bisector have
equal measures.
Lesson Practice 5B - sYsteMatic reVieW 5D
soLUtionsGeoMetrY 139
Lesson Practice 5BLesson Practice 5B1.
2.
3.
right
intersect
angle, linne segment
right
XZ
Follow the procedure in
4.
5.
6. the text.
Use a ruler to check that the line
ssegments of each side of the
bisector have equaal lengths.
Follow the procedure in the text.7.
Use a ruler to check that the line
segments off each side of the
bisector have equal lengths..
Follow the procedure in the text.
Use a pro
8.
ttractor to check that the angles
on each side off the bisector have
equal measures.
Follow th9. ee procedure in the text.
Use a protractor to chheck that the angles
on each side of the bisectoor have
equal measures.
Systematic Review 5CSystematic Review 5C1.
2.
3.
4.
5.
6.
7.
8.
c
f
b
e
d
a
e
180º
99.
10.
compass, straightedge
any angle between 1800º and 360º
null set
Use a ruler to check.
11.
12.
113. Use a ruler to check. the segment
on each siide of the bisector
should measure 114
U
.in
14. sse a protractor to check.
Use a protractor t15. oo check.
DeG and FeG
should each measure 34º.
∠ ∠
116.
17.
18.
19.
20.
reflex
right
acute
obtuse
straight
Systematic Review 5C1.
2.
3.
4.
5.
6.
7.
8.
c
f
b
e
d
a
e
180º
99.
10.
compass, straightedge
any angle between 1800º and 360º
null set
Use a ruler to check.
11.
12.
113. Use a ruler to check. the segment
on each siide of the bisector
should measure 114
U
.in
14. sse a protractor to check.
Use a protractor t15. oo check.
DeG and FeG
should each measure 34º.
∠ ∠
116.
17.
18.
19.
20.
reflex
right
acute
obtuse
straight
Systematic Review 5DSystematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
Systematic Review 5D1.
2.
3.
bisector
parallel
reflexx
perpendicular
similar
congruent
lines Qe
4.
5.
6.
7. ,, QD, eD
Q
yes: although this plan
8.
9.
10.
11.
Q
yes
ee is not
shown, any pair of intersecting
lines lie in the same plane.
a ruler to check12. Use ..
Use a ruler to check. the
segment on each
13.
sside of the
bisector should measure 112
.in
14. UUse a protractor to check.
Use a protractor 15. tto check.
aBG and cBG
should each measure 23º
∠ ∠
..
16.
17.
18.
19.
20.
sYsteMatic reVieW 5e - Lesson Practice 6B
soLUtions GeoMetrY140
Systematic Review 5ESystematic Review 5E1.
2.
3.
4.
5.
6.
7.
8.
f
e
b
c
g
a
d
falsee: Use a compass and
a straightedge
fa
9.
10.
true
llse: the two parts are
congruent.
false: the11. line will be
perpendicular only if
it forms a 990º angle.
Use a ruler to check.
Us
12.
13.
14.
true
ee a ruler to check.
the segment on each side off the
bisector should measure 78
Use a p
.in
15. rrotractor to check.
Use a protractor to chec16. kk.
XYG and ZYG
should each measure 10º.
∠ ∠
17. 24Q ++ =
+( ) = ( )+ =
− − = −
−
18 30
6 4 3 6 5
4 3 5
14 21 42
7 2
Y
Q Y
Q Y
Q D
Q
18.
++( ) = − ( )+ =
− =
−( ) = ( )− =
3 7 6
2 3 6
16 8 56
8 2 1 8 72 1
D
Q D
X
XX
19.
772 7 12 8
82
4
22 33 44
11 2 3 11 4
2
XX
X
X
X
X
= +=
= =
+ =
+( ) = ( )+
20.
33 42 4 32 1
12
== −=
=
XX
X
Systematic Review 5E1.
2.
3.
4.
5.
6.
7.
8.
f
e
b
c
g
a
d
falsee: Use a compass and
a straightedge
fa
9.
10.
true
llse: the two parts are
congruent.
false: the11. line will be
perpendicular only if
it forms a 990º angle.
Use a ruler to check.
Us
12.
13.
14.
true
ee a ruler to check.
the segment on each side off the
bisector should measure 78
Use a p
.in
15. rrotractor to check.
Use a protractor to chec16. kk.
XYG and ZYG
should each measure 10º.
∠ ∠
17. 24Q ++ =
+( ) = ( )+ =
− − = −
−
18 30
6 4 3 6 5
4 3 5
14 21 42
7 2
Y
Q Y
Q Y
Q D
Q
18.
++( ) = − ( )+ =
− =
−( ) = ( )− =
3 7 6
2 3 6
16 8 56
8 2 1 8 72 1
D
Q D
X
XX
19.
772 7 12 8
82
4
22 33 44
11 2 3 11 4
2
XX
X
X
X
X
= +=
= =
+ =
+( ) = ( )+
20.
33 42 4 32 1
12
== −=
=
XX
X
Lesson Practice 6ALesson Practice 6A1.
2.
3.
∠ ∠
∠ ∠
∠
aHG cHF
FHB GHD
aH
,
,
GG
GHD
LFK or JFH
cHa
HFK or JFL
DH
4.
5.
6.
7.
8.
∠
∠ ∠
∠
∠ ∠
∠ GG
9.
10.
40
65
º:
º:
vertical angles
vertical angles
111.
12.
90
50
º:
º:
supplementary angles
complementaary angles
115º: supplementary angles13.
14. 90º: vertical angles
15.
16.
17.
18.
19.
20.
f
a
e
b
d
c
Lesson Practice 6A1.
2.
3.
∠ ∠
∠ ∠
∠
aHG cHF
FHB GHD
aH
,
,
GG
GHD
LFK or JFH
cHa
HFK or JFL
DH
4.
5.
6.
7.
8.
∠
∠ ∠
∠
∠ ∠
∠ GG
9.
10.
40
65
º:
º:
vertical angles
vertical angles
111.
12.
90
50
º:
º:
supplementary angles
complementaary angles
115º: supplementary angles13.
14. 90º: vertical angles
15.
16.
17.
18.
19.
20.
f
a
e
b
d
c
Lesson Practice 6BLesson Practice 6B1.
2.
3.
∠ ∠
∠ ∠
∠
MnQ snr
MnQ tnP
Yr
,
,
ZZ
tnP
QnM or Pnr
tnP
YrZ or srn
sn
4.
5.
6.
7.
8.
∠
∠ ∠
∠
∠ ∠
∠ rr
9.
10.
55º: complementary angles
35º: vertical anngles
90º: supplementary angles
85º: suppl
11.
12. eementary angles
40º: vertical angles
55º:
13.
14. vvertical angles
alpha
complementary
supp
15.
16.
17. llementary
gamma
vertical
delta
18.
19.
20.
Lesson Practice 6B - sYsteMatic reVieW 6e
soLUtionsGeoMetrY 141
Lesson Practice 6B1.
2.
3.
∠ ∠
∠ ∠
∠
MnQ snr
MnQ tnP
Yr
,
,
ZZ
tnP
QnM or Pnr
tnP
YrZ or srn
sn
4.
5.
6.
7.
8.
∠
∠ ∠
∠
∠ ∠
∠ rr
9.
10.
55º: complementary angles
35º: vertical anngles
90º: supplementary angles
85º: suppl
11.
12. eementary angles
40º: vertical angles
55º:
13.
14. vvertical angles
alpha
complementary
supp
15.
16.
17. llementary
gamma
vertical
delta
18.
19.
20.
Systematic Review 6CSystematic Review 6C1. 2 5; : if the student referrred to
these angles using their three-letter
nnames, that would be correct
as well.
42.
3.
4
BFD
..
5.
6.
7.
BFe or aFD
BFD or aFc or aFe
º;
1
40 complemmentary angles
if ,8. 40 2 50 1 40º; º , ºm then m∠ = ∠ =
since 1 and 2 are
complementary. if m 1
∠ ∠
∠ = 40º,,
m 4 , since
1 and 4 are vertical a
then ∠ =
∠ ∠
40º
nngles.
supplementary angles
9.
10.
11.
1 4
140º;
or
aany two of angles 1, 2, and 4
Use
12.
13.
∠ ∠3; cFe
a ruler to check. the
segments on each side off the
bisector should measure 34
Use a p
.in
14. rrotractor to check. the
angles on each side of the
bisector should measure 26º.
perpendicu15. llar
90º
empty or null
16.
17.
18.
19.
20.
180
90
180
º
º
º
Systematic Review 6C1. 2 5; : if the student referrred to
these angles using their three-letter
nnames, that would be correct
as well.
42.
3.
4
BFD
..
5.
6.
7.
BFe or aFD
BFD or aFc or aFe
º;
1
40 complemmentary angles
if ,8. 40 2 50 1 40º; º , ºm then m∠ = ∠ =
since 1 and 2 are
complementary. if m 1
∠ ∠
∠ = 40º,,
m 4 , since
1 and 4 are vertical a
then ∠ =
∠ ∠
40º
nngles.
supplementary angles
9.
10.
11.
1 4
140º;
or
aany two of angles 1, 2, and 4
Use
12.
13.
∠ ∠3; cFe
a ruler to check. the
segments on each side off the
bisector should measure 34
Use a p
.in
14. rrotractor to check. the
angles on each side of the
bisector should measure 26º.
perpendicu15. llar
90º
empty or null
16.
17.
18.
19.
20.
180
90
180
º
º
º
Systematic Review 6DSystematic Review 6D1.
2.
true
false: they are compplementary.
true
false: Perpendicular angles
3.
4. were
not in the list of given information.
fa5. llse: ray GK is the common side.
true
39º: ve
6.
7. rrtical angles
51º: complementary angles
90º:
8.
9. perpendicular lines form
90º angles
right10.
11..
12.
13.
14.
15.
16.
17.
18.
1
supplementary
360º
f
e
b
a
g
d
99.
20.
h
c
Systematic Review 6D1.
2.
true
false: they are compplementary.
true
false: Perpendicular angles
3.
4. were
not in the list of given information.
fa5. llse: ray GK is the common side.
true
39º: ve
6.
7. rrtical angles
51º: complementary angles
90º:
8.
9. perpendicular lines form
90º angles
right10.
11..
12.
13.
14.
15.
16.
17.
18.
1
supplementary
360º
f
e
b
a
g
d
99.
20.
h
c
Systematic Review 6ESystematic Review 6E1.
2.
lines Qr, rV, and QV
rt, ,
º º
º
Xr Xt
3.
4.
360 8 45
90 90
√ =
∠ = ∠ =if m 1 , then m srV ºº
since they are supplementary.
srV is made up ∠ oof the three
smaller angles in the problem, so
tthe sum of their measures is
equal to that of ∠∠srV.
obtuse
yes: Both are 90º, so they add
5.
6.
uup to 180º.
no: complementary angles add
up t
7.
oo 90º.
if 's 2, 3 and 4 are congruent,
8.
9.
yes
∠
aand add up to 90º, the measure
of each must be 90º3
.
since 8 and 4 are vertical
ang
ºor 30
∠ ∠
lles, they are congruent,
so m 8 .
2: vert
∠ = 30º
10. iical angles
acute11.
12. m m m
m
∠ + ∠ + ∠ =
∠ = −
2 3 4 90
3 90 2
º
º 55 35
3 90 60 30
3
º º
º º º
+( )∠ = − =
∠ = ∠
m
m YrX m13. : vertical aangles
see #12
Use your
m YrX
ray rQ
∠ = ( )30º
14.
15. rruler to check that the
resulting line segmentss are equal
in length.
Use your protractor t16. oo check that
the resulting angles are equal
in measure.
17.
18.
−( ) = −( ) −( ) =
− ( ) = − ( )
72
7 7 49
152
15 15(( ) = −
− = − ( )( ) = −
− ( ) = − ( )( ) =
225
122 12 12 144
92
9 9
19.
20. −−81
sYsteMatic reVieW 6e - Lesson Practice 7B
soLUtions GeoMetrY142
Systematic Review 6E1.
2.
lines Qr, rV, and QV
rt, ,
º º
º
Xr Xt
3.
4.
360 8 45
90 90
√ =
∠ = ∠ =if m 1 , then m srV ºº
since they are supplementary.
srV is made up ∠ oof the three
smaller angles in the problem, so
tthe sum of their measures is
equal to that of ∠∠srV.
obtuse
yes: Both are 90º, so they add
5.
6.
uup to 180º.
no: complementary angles add
up t
7.
oo 90º.
if 's 2, 3 and 4 are congruent,
8.
9.
yes
∠
aand add up to 90º, the measure
of each must be 90º3
.
since 8 and 4 are vertical
ang
ºor 30
∠ ∠
lles, they are congruent,
so m 8 .
2: vert
∠ = 30º
10. iical angles
acute11.
12. m m m
m
∠ + ∠ + ∠ =
∠ = −
2 3 4 90
3 90 2
º
º 55 35
3 90 60 30
3
º º
º º º
+( )∠ = − =
∠ = ∠
m
m YrX m13. : vertical aangles
see #12
Use your
m YrX
ray rQ
∠ = ( )30º
14.
15. rruler to check that the
resulting line segmentss are equal
in length.
Use your protractor t16. oo check that
the resulting angles are equal
in measure.
17.
18.
−( ) = −( ) −( ) =
− ( ) = − ( )
72
7 7 49
152
15 15(( ) = −
− = − ( )( ) = −
− ( ) = − ( )( ) =
225
122 12 12 144
92
9 9
19.
20. −−81
Lesson Practice 7ALesson Practice 7A1.
2.
3.
transversal
exterior
interrior
congruent
alternate
parallel
same
co
4.
5.
6.
7.
8. nngruent
60º: vertical angles
60º: correspon
9.
10. dding angles
1 and 2 are supplementary, so
m
11. ∠ ∠
∠∠ = − ∠ = − =
∠ ∠
2
2 and 6 are corre
180 1 180 70 110º º º º.m
ssponding
angles, so they are congruent.
thus, mm 6
70º: corresponding angles
120º:
∠ = 110º.
12.
13. ccorresponding angles
120º: vertical angles14.
15.. yes: since 1 and 5 are
corresponding angles
∠ ∠
,, they have
the same measure. 's
5 and 17 are
∠
supplementary,
so angles 1 and 17 are also.
16..
17.
yes
no: they are alternate interior angles .
118. no: they are supplementary
angles and add upp to 180º. if
they were congruent,
they would booth be 90º.
yes: corresponding angles
it may
19.
hhelp to ignore line MP.
yes: angles 12 and
( )20. 13 are alternate
exterior angles.
it may help to ignore lines Lr and MP.( )
Lesson Practice 7A1.
2.
3.
transversal
exterior
interrior
congruent
alternate
parallel
same
co
4.
5.
6.
7.
8. nngruent
60º: vertical angles
60º: correspon
9.
10. dding angles
1 and 2 are supplementary, so
m
11. ∠ ∠
∠∠ = − ∠ = − =
∠ ∠
2
2 and 6 are corre
180 1 180 70 110º º º º.m
ssponding
angles, so they are congruent.
thus, mm 6
70º: corresponding angles
120º:
∠ = 110º.
12.
13. ccorresponding angles
120º: vertical angles14.
15.. yes: since 1 and 5 are
corresponding angles
∠ ∠
,, they have
the same measure. 's
5 and 17 are
∠
supplementary,
so angles 1 and 17 are also.
16..
17.
yes
no: they are alternate interior angles .
118. no: they are supplementary
angles and add upp to 180º. if
they were congruent,
they would booth be 90º.
yes: corresponding angles
it may
19.
hhelp to ignore line MP.
yes: angles 12 and
( )20. 13 are alternate
exterior angles.
it may help to ignore lines Lr and MP.( )
Lesson Practice 7BLesson Practice 7B1.
2.
3.
4.
false
true
true
false: thhey are always congruent.
false: two parallel5. lines are cut by
a transversal.
true
t
6.
7.
8.
true
rrue
110º: alternate interior angles
110º: c
9.
10. oorresponding angles
85º: corresponding angle11. ss
80º: corresponding angles
80º: alternate
12.
13. exterior angles
85º: vertical angles
yes:
14.
15. they add up to 180º.
yes it may help to ign16. oore line eF.
: they are supplementary an
( )17. no ggles.
yes: corresponding angles
no: they a
18.
19. rre corresponding angles,
but it is not stated thhat line ae line BF
since they are corre
|| .
20. ssponding angles,
if they are congruent, then linne ae line BF|| .
Lesson Practice 7B - sYsteMatic reVieW 7e
soLUtionsGeoMetrY 143
Lesson Practice 7B1.
2.
3.
4.
false
true
true
false: thhey are always congruent.
false: two parallel5. lines are cut by
a transversal.
true
t
6.
7.
8.
true
rrue
110º: alternate interior angles
110º: c
9.
10. oorresponding angles
85º: corresponding angle11. ss
80º: corresponding angles
80º: alternate
12.
13. exterior angles
85º: vertical angles
yes:
14.
15. they add up to 180º.
yes it may help to ign16. oore line eF.
: they are supplementary an
( )17. no ggles.
yes: corresponding angles
no: they a
18.
19. rre corresponding angles,
but it is not stated thhat line ae line BF
since they are corre
|| .
20. ssponding angles,
if they are congruent, then linne ae line BF|| .
Systematic Review 7CSystematic Review 7C1.
2.
∠ ∠ ∠ ∠
∠ ∠ ∠
3 4 5 6
1 2
; ; ;
; ; 77 8
3 6 5 4
1 8 2
;
;
;
∠
∠ ∠ ∠ ∠
∠ ∠ ∠
3.
4.
and and
and and
; ;
;
∠
∠ ∠ ∠ ∠
∠ ∠ ∠ ∠
7
1 5 3 7
2 6 4 8
5. and and
and and
66.
7.
8.
∠1
115º
they are alternate exterior angles..
hey are corresponding angles.
fal
9.
10.
11.
115º
t
sse: they are supplementary.
false: they do n12. oot lie on the
same line.
true
true: they a
13.
14. rre alternate
interior angles.
he two lines 15. t ccrossed by a
transversal are parallel.
hey 16. t aare perpendicular.
infinite
acute
obtuse
17.
18.
19.
220. reflex
Systematic Review 7DSystematic Review 7D1.
2.
3.
|| or is parallel to
7
33
44.
5.
6.
110º: they are supplementary.
supplementtary
vertical
if m then
m
7.
8. ∠ =
∠ = − =
7 72
8 180 72 1
º,
º º 008
8 108
6 108
º
º
º
if m∠ =
∠ =
, then
m : corresponding anggles
(other reasons why may also be ccorrect.)
hey are alternate exterior
9.
10.
110º
t aangles.
true
false: it is stated that line
11.
12. rs is not
parallel to line Vt.
true
true:
13.
14. vertical angles
orresponding angles are co15. c nngruent.
heir measures add up to 180º.
co
16.
17.
t
mmplementary
adjacent
beta
delta
18.
19.
20.
Systematic Review 7ESystematic Review 7E1. ∠ ∠1 3and
correspond
are
iing
and
corresponding
angles.
are
angle
∠ ∠3 11
ss.
1 so m 11=100º
: hey are alternat
∠ ≅ ∠ ∠11
1002. º t ee
exterior angles.
80º: 1 corresponds to 3,3. ∠ ∠ 3 and
4 are supplementary angles.
80º: they
∠
∠
4. are
supplementary angles.
13
: 7 a
5.
6.
7.
yes
yes ∠ nnd 2 are
alternate interior angles.
2 and 10
∠
∠ ∠ are
corresponding angles.
true:
8.
9.
10.
no
true
∠11 and 14 are alternate
exterior angles.
14 an
∠
∠ dd are
corresponding angles.
∠16
11.
12.
false
true:: Parallel lines do not
intersect.
he two l13. t iines cut by a
transversal are parallel.
hey14. t lie on the same plane.
gamma
alpha
15.
16.
17. 11−
= −−
=
−= −
=
1
31
3
12
12
11
1
18.
19.
20.
sYsteMatic reVieW 7e - sYsteMatic reVieW 8c
soLUtions GeoMetrY144
Systematic Review 7E1. ∠ ∠1 3and
correspond
are
iing
and
corresponding
angles.
are
angle
∠ ∠3 11
ss.
1 so m 11=100º
: hey are alternat
∠ ≅ ∠ ∠11
1002. º t ee
exterior angles.
80º: 1 corresponds to 3,3. ∠ ∠ 3 and
4 are supplementary angles.
80º: they
∠
∠
4. are
supplementary angles.
13
: 7 a
5.
6.
7.
yes
yes ∠ nnd 2 are
alternate interior angles.
2 and 10
∠
∠ ∠ are
corresponding angles.
true:
8.
9.
10.
no
true
∠11 and 14 are alternate
exterior angles.
14 an
∠
∠ dd are
corresponding angles.
∠16
11.
12.
false
true:: Parallel lines do not
intersect.
he two l13. t iines cut by a
transversal are parallel.
hey14. t lie on the same plane.
gamma
alpha
15.
16.
17. 11−
= −−
=
−= −
=
1
31
3
12
12
11
1
18.
19.
20.
Lesson Practice 8ALesson Practice 8A1.
2.
square (or rectangle)
rectaangle
triangle
rhombus (or quadrilateral)
t
3.
4.
5. rrapezoid
parallelogram (or quadrilateral)6.
7. P = 44 4 4 4 16
8 6 8 6 28
6 1 5 5 4 9 1
+ + + =
= + + + =
= + + =. . .
m
P in
P
8.
9. 66 5
10 10 10 10 40
3 6 5 7 8 24
. ft
. .
10.
11.
P cm
P
= + + + =
= + + + = 55
15 23 15 23 76
in
P mm
true
12.
13.
14.
15.
= + + + =
true
falsse: they add up to 360º.
true
false: a rig
16.
17. hht angle is possible
but not necessary.
18.
1
true
99.
20.
false: it has 2 pairs of parallel sides.
trrue
Lesson Practice 8A1.
2.
square (or rectangle)
rectaangle
triangle
rhombus (or quadrilateral)
t
3.
4.
5. rrapezoid
parallelogram (or quadrilateral)6.
7. P = 44 4 4 4 16
8 6 8 6 28
6 1 5 5 4 9 1
+ + + =
= + + + =
= + + =. . .
m
P in
P
8.
9. 66 5
10 10 10 10 40
3 6 5 7 8 24
. ft
. .
10.
11.
P cm
P
= + + + =
= + + + = 55
15 23 15 23 76
in
P mm
true
12.
13.
14.
15.
= + + + =
true
falsse: they add up to 360º.
true
false: a rig
16.
17. hht angle is possible
but not necessary.
18.
1
true
99.
20.
false: it has 2 pairs of parallel sides.
trrue
Lesson Practice 8BLesson Practice 8B1.
2.
3.
triangle
parallelogram
squuare
trapezoid
rhombus
rectangle
4.
5.
6.
7. P = + + +3 3 3 3 ==
= + + + =
= + + =
12
11 8 11 8 38
3 9 5 0 5 3 14 2. . . . f
m
P in
P
8.
9. tt
10.
11.
P in= + + + =18 32 45 23 118
length of unlabeled horizontal side:
length of unlabeled ve
4 2 2− = m
rrtical side:
length
6 4 2
4 4 2 2 2 6 20
− =
= + + + + + =
m
P m
12. of top horizontal side:
40 12 12 16
16
− − =
=
in
P ++ + + + + + +
=
12 12 16 40 16 12 12
136 in
13.
14.
triangle
quadrrilateral
square
rhombus
triangle
quad
15.
16.
17.
18. rrilateral
trapezoid
parallelogram
19.
20.
Systematic Review 8CSystematic Review 8C1.
2.
3.
4.
5.
6.
7.
b
a
f
d
e
c
P = + +3 5 7 ==
= + + + =
15
4 6 10 5 25
180
360
º
º
m
P in8.
9.
10.
11. is paralllel to, or ||
if two corresponding angles a12. rre
congruent, then the lines
are parallel.
i13. ff two lines are parallel,
corresponding angles
are congruent.
is parallel to, or ||
if a
14.
15. llternate exterior angles
are congruent, the twoo lines cut
by the transversal are parallel.
16..
17.
∠
∠ ∠
∠
12
106 3 4
4
º; and
and
are supplementary;
∠
∠
8
3
are corresponding.
as #17, or 18. same andd
and
∠
∠ ∠
6
6 8
are
alternate interior angles;
are supplementary.
are
co
19.
20.
74
3 11
º
∠ ∠and
rrresponding angles.
sYsteMatic reVieW 8c - sYsteMatic reVieW 8e
soLUtionsGeoMetrY 145
Systematic Review 8C1.
2.
3.
4.
5.
6.
7.
b
a
f
d
e
c
P = + +3 5 7 ==
= + + + =
15
4 6 10 5 25
180
360
º
º
m
P in8.
9.
10.
11. is paralllel to, or ||
if two corresponding angles a12. rre
congruent, then the lines
are parallel.
i13. ff two lines are parallel,
corresponding angles
are congruent.
is parallel to, or ||
if a
14.
15. llternate exterior angles
are congruent, the twoo lines cut
by the transversal are parallel.
16..
17.
∠
∠ ∠
∠
12
106 3 4
4
º; and
and
are supplementary;
∠
∠
8
3
are corresponding.
as #17, or 18. same andd
and
∠
∠ ∠
6
6 8
are
alternate interior angles;
are supplementary.
are
co
19.
20.
74
3 11
º
∠ ∠and
rrresponding angles.
Systematic Review 8DSystematic Review 8D1.
2.
3.
right
quadrilateral
squaare
rhombus
trapezoid
parallelogram
4.
5.
6.
7. P = + +5 7 111 23
4 6 10 5 25
10 10 10 10 40
=
= + + + =
= + + + =
m
P in
P cm
8.
9.
110. length of unlabeled
horizontal side:
10 2 3− − = 55
2 2 5 5 2 5
3 6 10 6 37
. .
in
P
in
= + + + +
+ + + =
11. transversal;; parallel
there may be alternate
explanations ffor #12, 13, 14.
are
alternate
12. 54º; ∠ ∠a and g
interior angles.
are
alternat
13. 54º; ∠ ∠b and d
ee interior angles.
14. 72 108º; º ,m d m g
theref
∠ + ∠ =
oore m º:∠ =2 108
alternate interior angles
acut15. ee
108º
supplementary or adjacent
FDe,
16.
17.
18. ∠ ∠FFGe, 3, or 2
a and b (or d and g)
if two
∠ ∠
19.
20. lines are perpendicular,
they form right angless.
Systematic Review 8D1.
2.
3.
right
quadrilateral
squaare
rhombus
trapezoid
parallelogram
4.
5.
6.
7. P = + +5 7 111 23
4 6 10 5 25
10 10 10 10 40
=
= + + + =
= + + + =
m
P in
P cm
8.
9.
110. length of unlabeled
horizontal side:
10 2 3− − = 55
2 2 5 5 2 5
3 6 10 6 37
. .
in
P
in
= + + + +
+ + + =
11. transversal;; parallel
there may be alternate
explanations ffor #12, 13, 14.
are
alternate
12. 54º; ∠ ∠a and g
interior angles.
are
alternat
13. 54º; ∠ ∠b and d
ee interior angles.
14. 72 108º; º ,m d m g
theref
∠ + ∠ =
oore m º:∠ =2 108
alternate interior angles
acut15. ee
108º
supplementary or adjacent
FDe,
16.
17.
18. ∠ ∠FFGe, 3, or 2
a and b (or d and g)
if two
∠ ∠
19.
20. lines are perpendicular,
they form right angless.
Systematic Review 8ESystematic Review 8E1.
2.
true
false: they add up tto 180º.
false: it has one pair
of para
3.
4.
true
lllel sides.
true
false
lengt
5.
6.
7.
8.
P in= + + =5 4 3 12
hh of unlabeled
horizontal side:
length
12 8 4− = in
of unlabeled vertical side:
8 2 6
8 6 4 2
− =
= + + + +
in
P 112 8 40+ = in
Qt or Qr or st or sQ or rt9.
every liine segment in the
drawing cuts through a pair
of parallel line segments.
or is perpend10. ⊥, iicular to
is parallel to
complement
11.
12.
||, or
aary
alternate
yes
no
13.
14.
15.
16.
17.
1
90 43 47º º º− =
no
88. if the midpoint of line segment
DP is point aa, a is the middle
point of the line segment.
199.
20.
slope
slope
= − =
=
2 4
1
;
;
y-intercept
y-interceppt = −2
sYsteMatic reVieW 8e - sYsteMatic reVieW 9c
soLUtions GeoMetrY146
Systematic Review 8E1.
2.
true
false: they add up tto 180º.
false: it has one pair
of para
3.
4.
true
lllel sides.
true
false
lengt
5.
6.
7.
8.
P in= + + =5 4 3 12
hh of unlabeled
horizontal side:
length
12 8 4− = in
of unlabeled vertical side:
8 2 6
8 6 4 2
− =
= + + + +
in
P 112 8 40+ = in
Qt or Qr or st or sQ or rt9.
every liine segment in the
drawing cuts through a pair
of parallel line segments.
or is perpend10. ⊥, iicular to
is parallel to
complement
11.
12.
||, or
aary
alternate
yes
no
13.
14.
15.
16.
17.
1
90 43 47º º º− =
no
88. if the midpoint of line segment
DP is point aa, a is the middle
point of the line segment.
199.
20.
slope
slope
= − =
=
2 4
1
;
;
y-intercept
y-interceppt = −2
Lesson Practice 9ALesson Practice 9A1. a bh= = ( ) ( )12 4 10 6 131 4 2. . . ft≈
22. a average= × =
+ ( ) = ( )
base height
( ) ( )10 152
5 252
5
== =
= = ( ) ( ) =
=
1252
62 5 2
12
12
19 11 104 5 2
. ft
.3.
4.
a bh m
a (( )9 2 11 82
7 4 77 7 2
8 6 48 2
. . . .+ ( ) =
= ( ) ( ) =
=
in
a in
a
5.
6. 112
4 6 12 2
6 6 36 2
5 6 7 82
3
( ) ( ) =
= ( ) ( ) =
= +
ft
. .
7.
8.
a m
a ( ) .. .
. .
5 23 45 2
12
5 4 3 10 75 2
67
( ) =
= ( ) ( ) =
= (
in
a cm
a
9.
10. )) ( ) =
= ( ) ( ) =
=
100 6 700 2
2 1 4 5 9 45 2
1
,
. . . ft
cm
a
a
11.
12.22
7 3 10 5 2( ) ( ) = . ft
13.
14.
15.
base, height
average
hallf
Lesson Practice 9BLesson Practice 9B1.
2.
a in
a
= ( ) ( ) =
=
7 4 4 75 35 15 2. . .
(( . . ) . .
. .
9 2 11 82
7 4 77 7 2
12
9 2 5 5
+ ( ) =
= ( ) (
in
a3. )) =
= + ( ) =
= +
25 3 2
12 162
8 112 2
9 1
.
( )
(
m
a in
a
4.
5. 332
3 33 2
12
3 3 5 5 9 075 2
( ) =
= ( ) ( ) =
) ft
. . . ft6.
7
a
..
8.
a m
a
= ( ) ( ) =
= + ( )
. . .
( )
05 05 0025 2
112 1562
70 ==
= ( ) ( ) =
=
9 380 2
12
5 33 3 5 9 3275 2
2 3
, ft
. . . ft
.
9.
10.
a
a 33 1 2 2 796 2
4 10 2 3 40 6 46
( ) ( ) =
= ( ) ( ) + ( ) ( ) = + =
. . in
a11. iin
a
2
12
28 12 168 212.
13.
14.
= ( ) ( ) = ft
perpendicular
trrapezoid
rectangle15.
Lesson Practice 9B1.
2.
a in
a
= ( ) ( ) =
=
7 4 4 75 35 15 2. . .
(( . . ) . .
. .
9 2 11 82
7 4 77 7 2
12
9 2 5 5
+ ( ) =
= ( ) (
in
a3. )) =
= + ( ) =
= +
25 3 2
12 162
8 112 2
9 1
.
( )
(
m
a in
a
4.
5. 332
3 33 2
12
3 3 5 5 9 075 2
( ) =
= ( ) ( ) =
) ft
. . . ft6.
7
a
..
8.
a m
a
= ( ) ( ) =
= + ( )
. . .
( )
05 05 0025 2
112 1562
70 ==
= ( ) ( ) =
=
9 380 2
12
5 33 3 5 9 3275 2
2 3
, ft
. . . ft
.
9.
10.
a
a 33 1 2 2 796 2
4 10 2 3 40 6 46
( ) ( ) =
= ( ) ( ) + ( ) ( ) = + =
. . in
a11. iin
a
2
12
28 12 168 212.
13.
14.
= ( ) ( ) = ft
perpendicular
trrapezoid
rectangle15.
Systematic Review 9CSystematic Review 9C1.
2.
a cm
a
= ( ) ( ) =
= +
12
5 6 15 2
13( 2212
12 204 2
7 6 42 2
1 5
( ) =
= ( ) ( ) =
=
)
ft
.
in
a
a
3.
4. (( ) ( ) =4 5 6 75 2. . in
5.
6.
rectangle, square
paralleloggram, rectangle,
square, rhombus
square
trap
7.
8. eezoid
a quadrilateral with two pairs
of paral
9.
llel sides
a quadrilateral with two pairs
of
10.
pparallel sides and four
congruent sides and fouur
right angles
yes: corresponding angles11.
12. yyes: alternate interior angles
yes: 5 be13. ∠ ≅ ∠7 ccause they are
corresponding angles, and
7∠ ≅ ∠155 because they are also
corresponding angles.
14..
15.
84
10
º: alternate interior angles
96º: m m∠ = ∠55
14 180 84 96
: alternate
interior angles. m∠ = − =º º ºº:
supplementary angles
: alternate inm m∠ = ∠11 14 tterior angles
(there are several wayss to find the answer.)
84º: : altern16. m m∠ = ∠5 10 aate
interior angles
: corresponding anm m∠ = ∠12 10 ggles
105º: :
corresponding angles
17. m m
m
∠ = ∠
∠
13 5
144 180 75= −º º:
supplementary angles
75º: corres18. pponding angles
75º: alternate exterior angle19. ss
length of base:
length of unlabele
20.
7 3 10+ = in
dd vertical side:
6 4 2
3 6 10 4 7 2 32
− =
= + + + + + =
in
P in
sYsteMatic reVieW 9c - sYsteMatic reVieW 9e
soLUtionsGeoMetrY 147
Systematic Review 9C1.
2.
a cm
a
= ( ) ( ) =
= +
12
5 6 15 2
13( 2212
12 204 2
7 6 42 2
1 5
( ) =
= ( ) ( ) =
=
)
ft
.
in
a
a
3.
4. (( ) ( ) =4 5 6 75 2. . in
5.
6.
rectangle, square
paralleloggram, rectangle,
square, rhombus
square
trap
7.
8. eezoid
a quadrilateral with two pairs
of paral
9.
llel sides
a quadrilateral with two pairs
of
10.
pparallel sides and four
congruent sides and fouur
right angles
yes: corresponding angles11.
12. yyes: alternate interior angles
yes: 5 be13. ∠ ≅ ∠7 ccause they are
corresponding angles, and
7∠ ≅ ∠155 because they are also
corresponding angles.
14..
15.
84
10
º: alternate interior angles
96º: m m∠ = ∠55
14 180 84 96
: alternate
interior angles. m∠ = − =º º ºº:
supplementary angles
: alternate inm m∠ = ∠11 14 tterior angles
(there are several wayss to find the answer.)
84º: : altern16. m m∠ = ∠5 10 aate
interior angles
: corresponding anm m∠ = ∠12 10 ggles
105º: :
corresponding angles
17. m m
m
∠ = ∠
∠
13 5
144 180 75= −º º:
supplementary angles
75º: corres18. pponding angles
75º: alternate exterior angle19. ss
length of base:
length of unlabele
20.
7 3 10+ = in
dd vertical side:
6 4 2
3 6 10 4 7 2 32
− =
= + + + + + =
in
P in
Systematic Review 9DSystematic Review 9D1.
2.
a cm
a
= ( ) ( ) =
=
12
8 8 32 2
6 5( . ++ ( ) =
= ( )( ) =
10 52
6 51 2
11 12 132 2
. )
ft
in
a
a
3.
4. == ( ) ( ) =5 1 2 2 11 22 2. . . in
5.
6.
7.
true
true
false: refleex angles measure
between 180º and 360º. a 175ºº
angle is obtuse.
false: it has 2 pairs of p8. aarallel sides.
if a parallelogram contains one
right angle, then it will contain
four right aangles, and will be a
rectangle, which is a speecial kind
of parallelogram.
false: th
9.
10.
true
eey add up to 360º.
yes: they are all 90º, be11. ccause
it is given that the lines
are perpendicuular.
79º: vertical angles
79º: correspond
12.
13. iing angles
supplementary
14. m∠ = − =3 180 79 101º º º:
aangles
yes15.
16.
17.
∠
∠ ∠ ∠ ∠
∠
16
6 9 5 10
7
; ;and and
;
;
and or and
and and
∠ ∠ ∠
∠ ∠ ∠ ∠
12 8 11
1 14 218. 113
4 15 3 16
;
;∠ ∠ ∠ ∠and or and
19.
20.
vertical
if aa quadrilateral is a trapezoid,
it has only onee pair
of parallel sides.
Systematic Review 9D1.
2.
a cm
a
= ( ) ( ) =
=
12
8 8 32 2
6 5( . ++ ( ) =
= ( )( ) =
10 52
6 51 2
11 12 132 2
. )
ft
in
a
a
3.
4. == ( ) ( ) =5 1 2 2 11 22 2. . . in
5.
6.
7.
true
true
false: refleex angles measure
between 180º and 360º. a 175ºº
angle is obtuse.
false: it has 2 pairs of p8. aarallel sides.
if a parallelogram contains one
right angle, then it will contain
four right aangles, and will be a
rectangle, which is a speecial kind
of parallelogram.
false: th
9.
10.
true
eey add up to 360º.
yes: they are all 90º, be11. ccause
it is given that the lines
are perpendicuular.
79º: vertical angles
79º: correspond
12.
13. iing angles
supplementary
14. m∠ = − =3 180 79 101º º º:
aangles
yes15.
16.
17.
∠
∠ ∠ ∠ ∠
∠
16
6 9 5 10
7
; ;and and
;
;
and or and
and and
∠ ∠ ∠
∠ ∠ ∠ ∠
12 8 11
1 14 218. 113
4 15 3 16
;
;∠ ∠ ∠ ∠and or and
19.
20.
vertical
if aa quadrilateral is a trapezoid,
it has only onee pair
of parallel sides.
Systematic Review 9ESystematic Review 9E1.
2.
a cm
P
= ( ) ( ) =
= +
( )12
8 4 16 2
8 77 8 5 20 8
7 132
3 30 2
5
. .
( )
+ =
= + ( ) =
= +
cm
a in
P
3.
4. 77 13 3 5 28 5+ + =. . in
5. length of unlabeled horizontaal
side: 14 5 5 4
3 5 3 4 14 5
15
− − =
= ( ) ( ) + ( ) ( ) + ( ) ( ) =
cm
a
++ + =
= + + + + + + +
=
12 70 97 2
14 8 4 3 5 3 5 8
50
90º
cm
P
cm
6.
7. : iit is given that MP is
perpendicular to Ln
↔
↔.
8.. m m m∠ + ∠ + ∠ =1 3 4 180º
because they are the three anngles
of a triangle. since m
m m
∠ =
∠ + ∠ +
4 90
1 3 90
º,
º ==
∠ + ∠ =
180
1 3 90
º
º.
,
or
bisector
perpendicul
m m
9.
10. aar bisector
find the average base
congruen
11.
12. tt
360º
check with ruler: line
segme
13.
14.
15.
360º
nnt should measure 3 12
check with ruler:
l
in
16.
iine segment on each side of the
bisector should measure 1 34
2 2 4 2 2
2 2
in
a X X X units
P X X
17. = ( )( ) =
= + ++ + =
= ( )( ) =
= + +
2 2 8
2 2 2 2
2
X X X units
a a a a units
P a a a
18.
++ =
= = ( )( ) =
2 6
12
12 2
12
a a units
a bh a B aB or aB uni19. tts
P a B c units
a X X X X
2
4 62
2 102
= + +
= + ( ) =( ) (20. ( )
= ( ) ( ) =
= ( ) + +( ) +
) 2
5 2 10 2 2
4 2 2
X
X X X units
P X X 66 2 1
14 3
X X
X units
( ) + +( )= +
sYsteMatic reVieW 9e - Lesson Practice 10B
soLUtions GeoMetrY148
Systematic Review 9E1.
2.
a cm
P
= ( ) ( ) =
= +
( )12
8 4 16 2
8 77 8 5 20 8
7 132
3 30 2
5
. .
( )
+ =
= + ( ) =
= +
cm
a in
P
3.
4. 77 13 3 5 28 5+ + =. . in
5. length of unlabeled horizontaal
side: 14 5 5 4
3 5 3 4 14 5
15
− − =
= ( ) ( ) + ( ) ( ) + ( ) ( ) =
cm
a
++ + =
= + + + + + + +
=
12 70 97 2
14 8 4 3 5 3 5 8
50
90º
cm
P
cm
6.
7. : iit is given that MP is
perpendicular to Ln
↔
↔.
8.. m m m∠ + ∠ + ∠ =1 3 4 180º
because they are the three anngles
of a triangle. since m
m m
∠ =
∠ + ∠ +
4 90
1 3 90
º,
º ==
∠ + ∠ =
180
1 3 90
º
º.
,
or
bisector
perpendicul
m m
9.
10. aar bisector
find the average base
congruen
11.
12. tt
360º
check with ruler: line
segme
13.
14.
15.
360º
nnt should measure 3 12
check with ruler:
l
in
16.
iine segment on each side of the
bisector should measure 1 34
2 2 4 2 2
2 2
in
a X X X units
P X X
17. = ( )( ) =
= + ++ + =
= ( )( ) =
= + +
2 2 8
2 2 2 2
2
X X X units
a a a a units
P a a a
18.
++ =
= = ( )( ) =
2 6
12
12 2
12
a a units
a bh a B aB or aB uni19. tts
P a B c units
a X X X X
2
4 62
2 102
= + +
= + ( ) =( ) (20. ( )
= ( ) ( ) =
= ( ) + +( ) +
) 2
5 2 10 2 2
4 2 2
X
X X X units
P X X 66 2 1
14 3
X X
X units
( ) + +( )= +
Lesson Practice 10ALesson Practice10A1.
2.
3.
isosceles
scalene
isoscelles
right
acute
obtuse
yes
he angles wou
4.
5.
6.
7.
8. t lld be 90º, 45º,
and 45º.
no:
yes:
9.
10.
5 7 15
8
+ <
+ 99 11>
11.
12.
13.
14.
isosceles
equilateral
angles
acutee
obtuse
scalene
right
riangles will
15.
16.
17.
18. t vvary.
ne angle must 90º.
riangles will var
o
t
=
19. yy.
ll angles must 90º.
riangles will vary.
a
t
<
20. two angles
must have the same measure.
Lesson Practice10A1.
2.
3.
isosceles
scalene
isoscelles
right
acute
obtuse
yes
he angles wou
4.
5.
6.
7.
8. t lld be 90º, 45º,
and 45º.
no:
yes:
9.
10.
5 7 15
8
+ <
+ 99 11>
11.
12.
13.
14.
isosceles
equilateral
angles
acutee
obtuse
scalene
right
riangles will
15.
16.
17.
18. t vvary.
ne angle must 90º.
riangles will var
o
t
=
19. yy.
ll angles must 90º.
riangles will vary.
a
t
<
20. two angles
must have the same measure.
Lesson Practice 10BLesson Practice10B1.
2.
3.
equilateral
scalene
isoscceles
obtuse
right
equiangular
no
in a ri
4.
5.
6.
7.
8. gght triangle, one angle is
90º, and the other ttwo must each
be < 90º.
yes:
yes:
9.
10.
10 11 12+ >
22 6 7+ >
11.
12.
13.
14.
15.
16.
two
three
90
90
must be
zero
less than
obtuse
riangles will v
10 8 18+ =
17.
18. t aary. all three angles
must have different measuures.
riangles will vary. one angle
must be
19. t
> 90º.
riangles will vary. angles must
hav
20. t
ee the same measure of 60º.
sYsteMatic reVieW 10c - sYsteMatic reVieW 10e
soLUtionsGeoMetrY 149
Systematic Review 10CSystematic Review 10C1.2.3.
isoscelesobtuseisosceeles, acutetriangle will have one 90º and tw
4.oo 45º angles
an equilateral triangle i5.6.
nono : ss also equiangular. since the angles are equal and add up to 180º, the measure
of each would be 180º3
=
= × × = × × =
=
60
12
3 12
2 23
12
72
83
5612
1
º.
7. a
443
4 23
2
05 05 0025 2
3 12
2 23
4
=
= ( ) ( ) =
= + +
. . .
in
a m
P
8.
9. 225
72
83
225
72
1515
83
1010
225
66
10530
8030
=
+ + =
× + × + × =
+ ++ =
=
= + + + =
13230
31730
10 1730
05 05 05 05 2. . . . .
in
P10. mmtrue11.
12.13.1
false: they add up to 180º.true
44.15.16.17.
truefalse: they are coplanar.truefallse: the Greek letter beta is falsetru
β.18.19. ee
length of unlabeled vertical sides:20.5 6 3 6. .− ==
+
2
1 7.
inlength of unlabeled horizontal side: 11 6 2 1 1 6 7
2 1 6 2 1 6 3 6 7
3
. . .
. . .
+ + =
= ( ) ( ) + ( ) ( ) + ( ) ( ) =
in
a
.. . . .2 3 2 25 2 31 6 2+ + = in
Systematic Review 10DSystematic Review 10D1.
2.
3.
equilateral
acute
scaleene; obtuse
triangles will vary. one angle
mu
4.
sst be greater than 90º, and
2 sides must be of equal length.
a right triangle may have
5.
6.
yes
sides
of three different lengths.
7. a = ( )12
12 16(( ) =
= + ( ) = ( ) ( ) =
= + +
96 2
4 122
3 8 3 24 2
16 20 1
cm
a m
P
8.
9.
22 48
5 4 5 12 26
=
= + + + =
cm
P m10.
11. yes: alternate inteerior angles
yes: alternate interior angles12.
133. m
m
m m m
m
∠ =
∠ =
∠ + ∠ + ∠ =
∠ + +
8 90
11 35
7 8 11 180
7 90 35
º
º
º
º º ==
∠ = − − =
180
7 180 90 35 55
º
º º º ºm
14. vertical; right (orr supplementary)
180º
yes
30º
15.
16.
17.
18.
19.
alpha
ccheck with a protractor. the two
smaller angless should each
measure 20.5º.
length of unlab20. eeled vertical sides:
length of unla
5 6 3 6 2. .− = in
bbeled
horizontal side:
1 7 1 6 2 1 1 6 7
7
. . . .+ + + =
= +
in
P 55 6 1 6 2 2 1
2 1 6 2 1 7 3 6 29 2
. . .
. . . .
+ + + +
+ + + + = in
Systematic Review 10ESystematic Review 10E1.
2.
3.
scalene
right
equlateraal, equiangular
all angles should be less tha4. nn
90º; no angles or sides should
have the same measure.
no
By definition, all angles in an
5.
6.
aacute triangle are less than 90º.
7. a = ( )12
6 18 4.(( ) =
= ( ) ( ) =
= + +
55 2 2
7 7 4 9 37 73 2
12 10 18
. ft
. . .8.
9.
a m
P .. . ft
. . . .
4 40 4
7 7 5 3 7 7 5 3 26
=
= + + + =
⊥ ∠
10.
11.
P m
; acD iis marked as a right
angle
bisects; m12.
1
∠ = ∠1 2m
33.
14.
acB and acD, or 1 and 2
c is the midpoint oof BD.
15.
16.
a
in
P
= ( )( ) − ( )( ) =
− =
=
11 11 7 7
121 49 72 2
111 11 11 11 44+ + + = in
For numbers 17-20,
the last terrm may vary.
17.
18.
19.
20.
Y X
Y X
Y X
Y
= − −
= +
= −
=
12
1
3 5
2
4XX + 3
sYsteMatic reVieW 10e - Lesson Practice 11B
soLUtions GeoMetrY150
Systematic Review 10E1.
2.
3.
scalene
right
equlateraal, equiangular
all angles should be less tha4. nn
90º; no angles or sides should
have the same measure.
no
By definition, all angles in an
5.
6.
aacute triangle are less than 90º.
7. a = ( )12
6 18 4.(( ) =
= ( ) ( ) =
= + +
55 2 2
7 7 4 9 37 73 2
12 10 18
. ft
. . .8.
9.
a m
P .. . ft
. . . .
4 40 4
7 7 5 3 7 7 5 3 26
=
= + + + =
⊥ ∠
10.
11.
P m
; acD iis marked as a right
angle
bisects; m12.
1
∠ = ∠1 2m
33.
14.
acB and acD, or 1 and 2
c is the midpoint oof BD.
15.
16.
a
in
P
= ( )( ) − ( )( ) =
− =
=
11 11 7 7
121 49 72 2
111 11 11 11 44+ + + = in
For numbers 17-20,
the last terrm may vary.
17.
18.
19.
20.
Y X
Y X
Y X
Y
= − −
= +
= −
=
12
1
3 5
2
4XX + 3
Lesson Practice 11ALesson Practice11A1.
2.
3.
4.
5.
6.
7.
8.
9.
c
d
b
e
f
a
5
6
1800 6 1 080
1 080 8 135
180 135 45
º , º
, º º
º º º
× =
=
− =
10.
11.
12.
÷
445 8 360
2 180
º º
º
× =
−( ) ( )13.
14.
n
dodecagon;
360º totaal ÷ 30º sides
decagon
=
+ =
−( )
12
8 2 10
2 180
15.
16.
;
n ºº º
º , º
, º
( ) => ( ) −( ) ( ) =
( ) =
15 2 180
13 180 2 340
2 340 117. ÷ 55 156
360 15 24
1
=
=
º
º º18. ÷
for each exterior angle;
880 24 156º º º− =
for each interior angle
Lesson Practice11A1.
2.
3.
4.
5.
6.
7.
8.
9.
c
d
b
e
f
a
5
6
1800 6 1 080
1 080 8 135
180 135 45
º , º
, º º
º º º
× =
=
− =
10.
11.
12.
÷
445 8 360
2 180
º º
º
× =
−( ) ( )13.
14.
n
dodecagon;
360º totaal ÷ 30º sides
decagon
=
+ =
−( )
12
8 2 10
2 180
15.
16.
;
n ºº º
º , º
, º
( ) => ( ) −( ) ( ) =
( ) =
15 2 180
13 180 2 340
2 340 117. ÷ 55 156
360 15 24
1
=
=
º
º º18. ÷
for each exterior angle;
880 24 156º º º− =
for each interior angle
Lesson Practice 11BLesson Practice11B1.
2.
3.
4.
5.
6.
7.
8.
9.
b
d
a
f
e
c
2
3
1800 3 540
540 5 108
180 108 72
72
º º
º º
º º º
º
× =
=
− =
×
10.
11.
12.
÷
55 360
2 180
360 36 10
=
−( ) ×
=
º
º
: º º
13.
14.
n
decagon ÷ sidees
triangles would mean 8 sides,
so it wo
15. six
uuld be an octagon.
16. n −( ) × => ( ) −( ) ×2 180 3 2 180º º ==
( ) × =
=
1 180
180
180 3 60
º
º
º º17.
18.
÷
exterior angles aadd up to 360º:
for each exterior a
360 3 120º º÷ =
nngle.
nterior angles
are 180º
i
− =120 60º º.
Lesson Practice11B1.
2.
3.
4.
5.
6.
7.
8.
9.
b
d
a
f
e
c
2
3
1800 3 540
540 5 108
180 108 72
72
º º
º º
º º º
º
× =
=
− =
×
10.
11.
12.
÷
55 360
2 180
360 36 10
=
−( ) ×
=
º
º
: º º
13.
14.
n
decagon ÷ sidees
triangles would mean 8 sides,
so it wo
15. six
uuld be an octagon.
16. n −( ) × => ( ) −( ) ×2 180 3 2 180º º ==
( ) × =
=
1 180
180
180 3 60
º
º
º º17.
18.
÷
exterior angles aadd up to 360º:
for each exterior a
360 3 120º º÷ =
nngle.
nterior angles
are 180º
i
− =120 60º º.
sYsteMatic reVieW 11c - sYsteMatic reVieW 11e
soLUtionsGeoMetrY 151
Systematic Review 11CSystematic Review 11C1.
2.
3.
4.
3
4
180 4 720
720
º º
º
× =
÷66 120
180 120 60
60 6 360
=
− =
× =
º
º º º
º º
5.
6.
7. square: exteerior angles
add up to 360º.
sides360 90 4º º÷ =
8. ffive
n
sides, so it would be
a pentagon
9. −( )2 1800 12 2 180
10 180 1 800
1 800 12 1
º º
º , º
, º
=> ( ) −( )= ( ) =
=10. ÷ 550
360 12 30
º
º ºcheck:
for each exterior angle.
÷ =
1180 30 150º º º− =
for each interior angle.
60º: 11. ∠∠
∠
acB is supplementary
to acD, which has a meassure
of 90º, so acB must also have a
measure
∠
oof 90º. acB, aBc
and Bac must add up to 180
∠ ∠
∠ ºº,
so m aBc
the ang
∠ = − +( ) =180 30 90 60º º º º .
:12. aBc lles add up to 90º.
, using reasoning13. ∠ =aDc 60º
similar to that used in question
number 11. siince aDc and
aDe are supplementary,
m aDe
∠
∠
∠ = 1800 120º º.
sup
− ∠ =m aDc
plementary14.
15.
1
equilateral
66.
17.
18.
19.
right
yes
yes: 9 8 15
1 2 1 1 1 4
+ >
=
( )( ) +
a
. . .(( )( ) + ( ) ( ) + ( )( ) =
+ + +
2 2 1 1 3 4 1 2 4 2
1 32 3 08 3 74 5
. . . . .
. . . .. .
. . . . . . . .
04 13 18 2
1 2 8 1 1 1 2 1 4 1 1 1 2 1
=
=
+ + + + + + +
m
P20.
11 2
9 1 2 18 2
( ) × =
× =. . m
Systematic Review 11DSystematic Review 11D1.
2.
3.
4.
4
5
180 5 900
900
º º
º
× =
÷77 128 57
180 128 57 51 43
51 43 7 360 0
≈ . º
º . º . º
. º .
5.
6.
− =
× = 11º
the .01º is due to rounding
in a previous steep.
hexagon: 7.
8.
9.
360 60 6
2
º º÷ =
−( )
sides
hexagon
n 1180 9 2 180
7 180 1 260
1 260 9 14
º º
º , º
, º
=> ( ) −( ) =
( ) =
=10. ÷ 00
360 9
º
º
check:
xterior angles add up to 360º.e
÷ = 440
180 40 140
º
º º º
for each exterior angle.
for
− =
eeach interior angle
GHK or FHJ
11.
12.
13.
JHK
yes : tthey are alternate interior
angles. it may helpp to extend JG
yes: they are alternate int
.→
14. eerior
angles. it may help to extend FK
iso
.→
15. ssceles
scalene
o: and the two shor
16.
17. n 1 1 2+ = , tt
sides need to add up to
something greater thaan the
long side.
check with a protra
18.
19.
a bh=
cctor:
angle should measure 125º
check with a20. protractor:
new angles should both
measure 62.55º
Systematic Review 11D1.
2.
3.
4.
4
5
180 5 900
900
º º
º
× =
÷77 128 57
180 128 57 51 43
51 43 7 360 0
≈ . º
º . º . º
. º .
5.
6.
− =
× = 11º
the .01º is due to rounding
in a previous steep.
hexagon: 7.
8.
9.
360 60 6
2
º º÷ =
−( )
sides
hexagon
n 1180 9 2 180
7 180 1 260
1 260 9 14
º º
º , º
, º
=> ( ) −( ) =
( ) =
=10. ÷ 00
360 9
º
º
check:
xterior angles add up to 360º.e
÷ = 440
180 40 140
º
º º º
for each exterior angle.
for
− =
eeach interior angle
GHK or FHJ
11.
12.
13.
JHK
yes : tthey are alternate interior
angles. it may helpp to extend JG
yes: they are alternate int
.→
14. eerior
angles. it may help to extend FK
iso
.→
15. ssceles
scalene
o: and the two shor
16.
17. n 1 1 2+ = , tt
sides need to add up to
something greater thaan the
long side.
check with a protra
18.
19.
a bh=
cctor:
angle should measure 125º
check with a20. protractor:
new angles should both
measure 62.55º
Systematic Review 11ESystematic Review 11E1.
2.
3.
4.
7
8
180 8 1 440
1 4
º , º
,
× =
440 10 144
180 144 36
36 10 360
÷ =
− =
× =
º
º º º
º º
5.
6.
7. trianglle: sides360 120 3
2 180 20
º º
º
÷ =
−( ) => (8.
9.
octagon
n )) −( ) =
( ) =
=
2 180
18 180 3 240
3 240 20 162
º
º , º
, º º10. ÷
checkk:
85º: vertical an
360 20 18
180 18 162
º º
º º º
÷ =
− =
11. ggles
supplementary angles
12.
13.
180 85 95º º º:− =
∠m JJFK
m GJK m
= − +( ) =
− =
∠ = −
180 85 45
180 130 50
90
º º º
º º º
º14. ∠∠ =
− =
∠
FJG
90 45 45º º º
the measure of is
unnecessa
α
rry for solving
this question.
average bas15. a = ee height
×
= + × = × =
=
=
a
m
P
10 172
6 272
61
1622
81 2
616. ++ + + =
= −− + = −
10 11 17 44
11
(
m
Y XX Y
17.or
multiplying bboth sides by 1)−
− =
+ + =+ = −
= − −
X Y
X YY X
Y X
1
2 4 04 2
2 4
18.
119.
20.
Y XX Y orX Y
X YY X
Y
= +− + =
− = −
+ − =− = −
4 24 24 2
2 8 02 8
2 == − +
= − +
X
Y X
812
4
sYsteMatic reVieW 11e - Lesson Practice 12B
soLUtions GeoMetrY152
Systematic Review 11E1.
2.
3.
4.
7
8
180 8 1 440
1 4
º , º
,
× =
440 10 144
180 144 36
36 10 360
÷ =
− =
× =
º
º º º
º º
5.
6.
7. trianglle: sides360 120 3
2 180 20
º º
º
÷ =
−( ) => (8.
9.
octagon
n )) −( ) =
( ) =
=
2 180
18 180 3 240
3 240 20 162
º
º , º
, º º10. ÷
checkk:
85º: vertical an
360 20 18
180 18 162
º º
º º º
÷ =
− =
11. ggles
supplementary angles
12.
13.
180 85 95º º º:− =
∠m JJFK
m GJK m
= − +( ) =
− =
∠ = −
180 85 45
180 130 50
90
º º º
º º º
º14. ∠∠ =
− =
∠
FJG
90 45 45º º º
the measure of is
unnecessa
α
rry for solving
this question.
average bas15. a = ee height
×
= + × = × =
=
=
a
m
P
10 172
6 272
61
1622
81 2
616. ++ + + =
= −− + = −
10 11 17 44
11
(
m
Y XX Y
17.or
multiplying bboth sides by 1)−
− =
+ + =+ = −
= − −
X Y
X YY X
Y X
1
2 4 04 2
2 4
18.
119.
20.
Y XX Y orX Y
X YY X
Y
= +− + =
− = −
+ − =− = −
4 24 24 2
2 8 02 8
2 == − +
= − +
X
Y X
812
4
Lesson Practice 12ALesson Practice12A1.
2.
3.
sphere
circumference
chorrd
radius
diameter
Ge
secto
4.
5.
6.
7.
, , ,Gc Ga or GD
rr
arc
tangent
ellipse
perpendicular
se
8.
9.
10.
11.
12. ccant
4
86º: the measure of
13.
14.
15.
360 60 300º º º− =
an
intercepted arc is the same as
the measure of the central
angle that intercepts it.
16. 86ºº º:÷2 43= the measure of an
inscribed angle is hhalf the
measure of a central angle
interceptingg the same arc.
100º: answers that are close17.
are acceptable.
100º: answers that are clos18. ee are
acceptable, but the answers to
17 and 18 must be the same.
Lesson Practice 12BLesson Practice12B1.
2.
3.
circumference
chord
spherre
radius
radius
diameter
tangent
arc
se
4.
5.
6.
7.
8.
9. cctor
two
one
ellipse
10.
11.
12.
13.
14.
360 270 90
3
º º º− =
115.
16.
17.
18.
44
442
22
90
90
º
º º
º
º
=
Lesson Practice 12B - sYsteMatic reVieW 12D
soLUtionsGeoMetrY 153
Lesson Practice12B1.
2.
3.
circumference
chord
spherre
radius
radius
diameter
tangent
arc
se
4.
5.
6.
7.
8.
9. cctor
two
one
ellipse
10.
11.
12.
13.
14.
360 270 90
3
º º º− =
115.
16.
17.
18.
44
442
22
90
90
º
º º
º
º
=
Systematic Review 12CSystematic Review 12C1.
2.
3.
4.
cB or cD
tangent
aB
seecant
sphere
ellipse
circumference
5
the
5.
6.
7.
8.
9. mmeasure of an inscribed
angle is half the measuure of the
arc that it intercepts, so it would
be
check with a ruler and protra
10 2 20º º.× =
10. cctor
answers that are close are
accepta
11. 44º:
bble.
right:
1 and 2 are complementary.
y
12.
13.
∠ ∠
ees: nLP and are
alternate interior angl
∠ ≅ ∠MPL,
ees.
supplementary angles
14.
15.
m m∠ = − ∠ =7 180 5 112º º
mm
m
∠ =
∠ = ∠ =
7 112
7 112
º
º
;
m rMn
alternate interior anglles
octagon:
quadrilatera
16.
17.
360 45 8º º÷ = sides
ll: any answer
naming a specific kind of
quadrillateral is acceptable.
18. n −( ) => ( ) −( )2 180 7 2 180º ºº
º º
º . º
º
=
( ) =5 180 900
900 7 128 57
360 7
19.
20.
÷ ≈
÷ sides ≈≈ 51 43
180 51 43 128 57
. º
º . º . º
per
exterior angle
− =
pper interior angle
Systematic Review 12C1.
2.
3.
4.
cB or cD
tangent
aB
seecant
sphere
ellipse
circumference
5
the
5.
6.
7.
8.
9. mmeasure of an inscribed
angle is half the measuure of the
arc that it intercepts, so it would
be
check with a ruler and protra
10 2 20º º.× =
10. cctor
answers that are close are
accepta
11. 44º:
bble.
right:
1 and 2 are complementary.
y
12.
13.
∠ ∠
ees: nLP and are
alternate interior angl
∠ ≅ ∠MPL,
ees.
supplementary angles
14.
15.
m m∠ = − ∠ =7 180 5 112º º
mm
m
∠ =
∠ = ∠ =
7 112
7 112
º
º
;
m rMn
alternate interior anglles
octagon:
quadrilatera
16.
17.
360 45 8º º÷ = sides
ll: any answer
naming a specific kind of
quadrillateral is acceptable.
18. n −( ) => ( ) −( )2 180 7 2 180º ºº
º º
º . º
º
=
( ) =5 180 900
900 7 128 57
360 7
19.
20.
÷ ≈
÷ sides ≈≈ 51 43
180 51 43 128 57
. º
º . º . º
per
exterior angle
− =
pper interior angle
Systematic Review 12DSystematic Review 12D1.
2.
3.
diameter
diameter
radiuus
secant
three
ellipses
rectangle, square
4.
5.
6.
7. ,, rhombus,
parallelogram
circumference
inscri
8.
9. bbed
PLM or MLP
10.
11.
12.
13
35 2 70
360 70 290
º º
º º º
× =
− =
..
14.
vertical angles
givenm m
m
∠ + ∠ = ( )∠ = −
1 2 90
1 90
º
º 558 32
5
º º=
∠( )m is unnecessary information
For 15. tthis problem it may be helpful
to ignore everytthing except
LMn. the measures of the
angles iin this triangle must add
up to 180º:
m nLM m∠ = ∠22 1
58 32 90
3 180 5
180 90
+ ∠ =
+ =
∠ = − ∠ + ∠( ) =
−
m
m m nLM m
º º º
º
º º ++( ) =
− =
68
180 158 22
º
º º º
16. line segment, line, or rray
obtuse angle
rhombus
scalene triangl
17.
18.
19. ee
octagon20.
sYsteMatic reVieW 12e - Lesson Practice 13B
soLUtions GeoMetrY154
Systematic Review 12ESystematic Review 12E1.
2.
3.
4.
ellipse
chord
radius
ddiameter or chord
arc
sector
12perpend
5.
6.
7.
8.
9.
a
iicular
heck your drawing using a ruler
and a
10. c
protractor.
225º: answers that are close to11. this
are acceptable.
512.
13.
14.
15
6
180 6 1 080º , º× =
..
16.
17.
18.
1 080 8 135
180 135 45
45 8 360
, º º
º º º
º º
÷ =
− =
× =
YY X Y X
Y X X X
XXX
− = => = +
+ = − => +( ) + = −
+ = −= −
2 4 2 4
5 2 4 5
3 4 53 9
== −
+ = − => + −( ) = −
= −
= − −( )−
3
5 3 5
2
3 2
Y X Y
Y
Y
solution ,
19. 44 4 4 4
2 2 4 4 2 2
6 4 26 6
X Y X
Y X X X
XXX
= => = +
+ = − => +( ) + = −
+ = −= −== −
− = => − −( ) =
+ ==
= −( )
1
4 4 4 1 4
4 40
1 0
Y X Y
YY
solution ,
20.. Y X Y X
Y X X X
X
X
Y X
− = => =
− = − => ( ) − = −
− = −
= −−
=
−
0
3 6 3 6
2 662
3
== => − ( ) =
=
= ( )
0 3 0
3
3 3
Y
Y
solution ,
Systematic Review 12E1.
2.
3.
4.
ellipse
chord
radius
ddiameter or chord
arc
sector
12perpend
5.
6.
7.
8.
9.
a
iicular
heck your drawing using a ruler
and a
10. c
protractor.
225º: answers that are close to11. this
are acceptable.
512.
13.
14.
15
6
180 6 1 080º , º× =
..
16.
17.
18.
1 080 8 135
180 135 45
45 8 360
, º º
º º º
º º
÷ =
− =
× =
YY X Y X
Y X X X
XXX
− = => = +
+ = − => +( ) + = −
+ = −= −
2 4 2 4
5 2 4 5
3 4 53 9
== −
+ = − => + −( ) = −
= −
= − −( )−
3
5 3 5
2
3 2
Y X Y
Y
Y
solution ,
19. 44 4 4 4
2 2 4 4 2 2
6 4 26 6
X Y X
Y X X X
XXX
= => = +
+ = − => +( ) + = −
+ = −= −== −
− = => − −( ) =
+ ==
= −( )
1
4 4 4 1 4
4 40
1 0
Y X Y
YY
solution ,
20.. Y X Y X
Y X X X
X
X
Y X
− = => =
− = − => ( ) − = −
− = −
= −−
=
−
0
3 6 3 6
2 662
3
== => − ( ) =
=
= ( )
0 3 0
3
3 3
Y
Y
solution ,
Lesson Practice 13ALesson Practice13A1.
2.
3.
radius
c
circumference
= πdd or c r
a r
x y, ,
=
=
2
2
π
π
π
4.
5. or short axis, long axis,
latitude
longitude
minut
π( )6.
7.
8.
9.
square
ees
rime meridian10.
11.
p
c r= ( ) ( ) ( ) =2 2 3 14 3 18 84π ≈ . .
. .
in
a r in
a r
12.
13.
= ( ) ( ) =
=
π ≈
π ≈
2 3 14 32 28 26 2
2 227
72(( ) = ( ) =
= ( ) × ( ) ×
( ) ( )
227
49 154 2
12
12 12
8
6 4 3 1.
m
a14. π ≈
44 75 36 2
50 7 8 41
18 58 7
( ) = . ft
º ' ; º '
º ' ;
15.
16.
n e
n 22 50
4 082
4 082 1 6 6 531 2
º '
,
, . , .
e
mi
km
17.
18. × =
Lesson Practice 13BLesson Practice13B1.
2.
3.
diameter
circumference
arrea
degrees
latitude
4.
5.
6.
7.
8.
length
longitude0;
llatitude; longitude
seconds
square
9.
10.
11. c d= π ≈ 3.. .
. .
14 10 31 4
2 3 14 52 78 5 2
( ) ( ) =
= ( ) ( ) =
in
a r in12.
1
π ≈
33.
14.
c d m
a
= × =
= ( ) × ( ) ×
( ) ( )
π ≈
π ≈
227
141
44
12
10 12
6
5 3 3.. . ft
º ' ; º '
º '
14 47 1 2
40 43 74 01
33 55
( ) =
15.
16.
n W
s;; º '
,
, . , .
18 22
7 804
7 804 1 6 12 486 4
e
mi
km
17.
18. × =
Lesson Practice 13B - sYsteMatic reVieW 13D
soLUtionsGeoMetrY 155
Lesson Practice13B1.
2.
3.
diameter
circumference
arrea
degrees
latitude
4.
5.
6.
7.
8.
length
longitude0;
llatitude; longitude
seconds
square
9.
10.
11. c d= π ≈ 3.. .
. .
14 10 31 4
2 3 14 52 78 5 2
( ) ( ) =
= ( ) ( ) =
in
a r in12.
1
π ≈
33.
14.
c d m
a
= × =
= ( ) × ( ) ×
( ) ( )
π ≈
π ≈
227
141
44
12
10 12
6
5 3 3.. . ft
º ' ; º '
º '
14 47 1 2
40 43 74 01
33 55
( ) =
15.
16.
n W
s;; º '
,
, . , .
18 22
7 804
7 804 1 6 12 486 4
e
mi
km
17.
18. × =
Systematic Review 13CSystematic Review 13C1. a r= =π ≈2 22
772
2 3( )( ) 88 5 2. in
2.
3.
trapezoid
check with a ruler: diameteer
should be 7.5 in
c r= ( ) ( ) ( ) =2 2 3 14 3 75 23 55π ≈ . . . in
a
4.
5.
6.
7.
radius
latitude
longitude
= ( ) ×12
14 12
4(( ) ×
( )( ) ( ) =
π ≈
7 2 3 14 43 96 2
4
. . in
8. : a regular paralllelogram
is a square.
se a ruler and protra9. U cctor
to check.
320º: it may be easier to mea10. ssure
the acute angle, and subtract that
number from 360º.
11.
12.
64 09 21 57
39 55 1
º ' ; º '
º ' ;
n W
n 116 23
4 905
4 905 1 6 7 848
º '
,
, . ,
e
mi
km
13.
14.
15.
× =
eGDD; vertical angles
complementary: they are f16. oormed
from perpendicular lines.
obtuse
stra
17.
18. iight
triang
19.
20.
c r units= ( ) ( ) ( ) =2 2 3 14 8 50 24π ≈ . .
lle:
semici
. . . . fta bh= = +( ) ( ) =12
12
3 3 3 3 5 5 18 15 2
rrcle:
(half the area of the whole circla r= 12
2π ee)
total:
a
a
≈ ≈12
3 14 3 32 17 1 2
18 15 17 1
. . . ft
. .
( ) ( )
= + = 335 25 2. ft
Systematic Review 13C1. a r= =π ≈2 22
772
2 3( )( ) 88 5 2. in
2.
3.
trapezoid
check with a ruler: diameteer
should be 7.5 in
c r= ( ) ( ) ( ) =2 2 3 14 3 75 23 55π ≈ . . . in
a
4.
5.
6.
7.
radius
latitude
longitude
= ( ) ×12
14 12
4(( ) ×
( )( ) ( ) =
π ≈
7 2 3 14 43 96 2
4
. . in
8. : a regular paralllelogram
is a square.
se a ruler and protra9. U cctor
to check.
320º: it may be easier to mea10. ssure
the acute angle, and subtract that
number from 360º.
11.
12.
64 09 21 57
39 55 1
º ' ; º '
º ' ;
n W
n 116 23
4 905
4 905 1 6 7 848
º '
,
, . ,
e
mi
km
13.
14.
15.
× =
eGDD; vertical angles
complementary: they are f16. oormed
from perpendicular lines.
obtuse
stra
17.
18. iight
triang
19.
20.
c r units= ( ) ( ) ( ) =2 2 3 14 8 50 24π ≈ . .
lle:
semici
. . . . fta bh= = +( ) ( ) =12
12
3 3 3 3 5 5 18 15 2
rrcle:
(half the area of the whole circla r= 12
2π ee)
total:
a
a
≈ ≈12
3 14 3 32 17 1 2
18 15 17 1
. . . ft
. .
( ) ( )
= + = 335 25 2. ft
Systematic Review 13DSystematic Review 13D1. c r x x
x
= =2 2 227
3 5
2 227
π ≈
.
xx in
diameter
72
22=
2.
3.
4.
square or rectangle
tangeent
latitude
longitude
5.
6.
7. a = ( ) × ( ) ×
( )
12
6 12
2
3 1
π ≈
(( ) ( ) =
−( ) => ( ) −( )
3 14 9 42 2
2 180 20 2
. .
º
in
n
8.
9.
secant
1180
18 180 3 240
3 240 20 162
180 162
º
º , º
, º º
º º
=
( ) =
=
−
÷
10. ==
=
( ) ( ) =
−
18
360 20 18
2 50 100
360 100
º
º º
º º
º º
or
÷
11.
12. == 260
100
34 36 58 27
37 49
º
º
º ' ; º '
º º ;
13.
14.
15.
s W
s 1144 58º ' e
FGe BGc
m BGc m
16. ∠ ≅ ∠
∠ + ∠
: vertical angles
aaGB
m aGB
=
∠ = − =
90
90 43 47
º
º º º
:
complementary angles
117. m eGD
m eG
∠ = − =
∠
90 43 47º º º:
complementary angles
cc
m BGc m FGc m FGB
= + =
∠ = ∠ − ∠ =
− =
90 47 137
180 135 4
º º º
º º
18.
55
90
135 90 45
4
º
º
º º º
19. m aGB m FGB
m eGD m aGB
∠ = ∠ − =
− =
∠ = ∠ = 55
12
2
12
2
º:
vertical angles
semicircle:20.
c r= ( )
(
π ≈
)) ( ) ( ) =
+
3 14 3 3 10 36
6 2 6 2
. . . ft
. .
sides of triangle:
==
+ =
12 4
10 36 12 4 22 76
. ft
:
. . . ft
total
sYsteMatic reVieW 13D - Lesson Practice 14a
soLUtions GeoMetrY156
Systematic Review 13D1. c r x x
x
= =2 2 227
3 5
2 227
π ≈
.
xx in
diameter
72
22=
2.
3.
4.
square or rectangle
tangeent
latitude
longitude
5.
6.
7. a = ( ) × ( ) ×
( )
12
6 12
2
3 1
π ≈
(( ) ( ) =
−( ) => ( ) −( )
3 14 9 42 2
2 180 20 2
. .
º
in
n
8.
9.
secant
1180
18 180 3 240
3 240 20 162
180 162
º
º , º
, º º
º º
=
( ) =
=
−
÷
10. ==
=
( ) ( ) =
−
18
360 20 18
2 50 100
360 100
º
º º
º º
º º
or
÷
11.
12. == 260
100
34 36 58 27
37 49
º
º
º ' ; º '
º º ;
13.
14.
15.
s W
s 1144 58º ' e
FGe BGc
m BGc m
16. ∠ ≅ ∠
∠ + ∠
: vertical angles
aaGB
m aGB
=
∠ = − =
90
90 43 47
º
º º º
:
complementary angles
117. m eGD
m eG
∠ = − =
∠
90 43 47º º º:
complementary angles
cc
m BGc m FGc m FGB
= + =
∠ = ∠ − ∠ =
− =
90 47 137
180 135 4
º º º
º º
18.
55
90
135 90 45
4
º
º
º º º
19. m aGB m FGB
m eGD m aGB
∠ = ∠ − =
− =
∠ = ∠ = 55
12
2
12
2
º:
vertical angles
semicircle:20.
c r= ( )
(
π ≈
)) ( ) ( ) =
+
3 14 3 3 10 36
6 2 6 2
. . . ft
. .
sides of triangle:
==
+ =
12 4
10 36 12 4 22 76
. ft
:
. . . ft
total
Systematic Review 13ESystematic Review 13E1.
2.
3.
4.
diameter
227
a r= π 2
3.114
12
9 12
3
4 5
5.
6.
7.
latitude
longitude
a = ( ) × ( ) ×
( )
π ≈
. 11 5 3 14 21 2 2. . .( ) ( ) = in
8.
9.
10.
11.
60
sphere
diameter
aa r m
c d
= ( ) ( ) =
= ( ) (π ≈
π ≈
2 3 14 3 52
38 47 2
3 14 1 25
. . .
. .12. )) =
=
3 925 3 93
360 40 9
. .
º º
or cm
13.
14.
15.
÷
infinite
exxterior
alternate exterior
MLK, ceH
16.
17.
18.
1
GHe
99.
3 2 12
2 4 5 8 2 10
11 22
Y X
Y X Y X
Y
+ =
( ) − =( ) => − =
=
Y
Y X
=
+ = => ( )2
3 2 12 3 2 ++ =
+ ===
= ( )− = −
−
2 12
6 2 122 6
3
3 2
3
X
XXX
Y X
Y
solution ,
20.
22 4 2 4
1
3 1 3
2
X Y X
X
Y X Y
Y
= => − + =
=
− = => − ( ) = −
= −
=
–
–
solution 11 2, −( )
Systematic Review 13E1.
2.
3.
4.
diameter
227
a r= π 2
3.114
12
9 12
3
4 5
5.
6.
7.
latitude
longitude
a = ( ) × ( ) ×
( )
π ≈
. 11 5 3 14 21 2 2. . .( ) ( ) = in
8.
9.
10.
11.
60
sphere
diameter
aa r m
c d
= ( ) ( ) =
= ( ) (π ≈
π ≈
2 3 14 3 52
38 47 2
3 14 1 25
. . .
. .12. )) =
=
3 925 3 93
360 40 9
. .
º º
or cm
13.
14.
15.
÷
infinite
exxterior
alternate exterior
MLK, ceH
16.
17.
18.
1
GHe
99.
3 2 12
2 4 5 8 2 10
11 22
Y X
Y X Y X
Y
+ =
( ) − =( ) => − =
=
Y
Y X
=
+ = => ( )2
3 2 12 3 2 ++ =
+ ===
= ( )− = −
−
2 12
6 2 122 6
3
3 2
3
X
XXX
Y X
Y
solution ,
20.
22 4 2 4
1
3 1 3
2
X Y X
X
Y X Y
Y
= => − + =
=
− = => − ( ) = −
= −
=
–
–
solution 11 2, −( )
Lesson Practice 14ALesson Practice14A1.
2.
3.
base height,
faces
squarees
edges
circle
cubic
vertices
4.
5.
6.
7.
8. V = ( )( )(4 4 4)) =
= ( ) ( ) ( ) =
= =
(
64 3
5 4 3 60 3
2
3 14
ft
.
in
V
V Bh r h
9.
10. π ≈
)) ( ) ( ) =
= ( ) ( )( ) =
=
102 5 1 570 3
5 7 4 140 3
, ft
ft11.
12.
V
V BBh r h=
( ) ( ) ( ) =
π ≈2
3 14 102 25 7 850 3. , ft
Lesson Practice 14B - sYsteMatic reVieW 14D
soLUtionsGeoMetrY 157
Lesson Practice 14BLesson Practice14B1.
2.
3.
4.
5.
6.
7.
8.
9.
g
a
c
b
d
h
f
e
V = 220 30 10 6 000 3
2
3 14 152
( ) ( ) ( ) =
= =
( ) ( )
, ft
.
10. V Bh r hπ ≈
660 42 390 3
10 10 10 1 000 3
( ) =
= ( ) ( ) ( ) =
,
, ft
ft
V
V
11.
12. == =
( ) ( ) ( ) =
Bh r hπ ≈2
3 14 102 15 4 710 3. , ft
Systematic Review 14CSystematic Review 14C1.
2.
V m
V Bh
= ( ) ( ) ( ) =
= =
3 5 4 60 3
ππ ≈
π ≈
r h
V Bh r h
2
3 14 42 4 200 96 3
2
3 14
. . ft
.
( ) ( )( ) =
= =
(3.
)) ( ) ( ) =
= ( ) ( ) =
=
62 10 1 130 4 3
8 8 64 2
2
, . ft
4.
5.
a cm
a rπ ≈ 33 14 62 113 04 2
3 5 3 5 3 14
. .
.
( ) ( ) =
= ( ) ( ) ( ) ( ) ( ) (cm
a6. π ≈ )) =
−( ) => ( ) −( ) =
( ) ( ) =
47 1 2
2 180 8 2 180
6 180 1
.
º º
º
cm
n7.
,, º
, º º
º º
080
1 080 8 135
180 135
total
÷ =
− =
per angle
8. 445 360 8 45
45 8 360
360 10 36
º º º;
º º
º º
or
sid
÷
÷
=
( ) =
=9. ees
c r in
10.
11.
12.
sphere
= ( ) ( ) ( ) =2 2 3 14 6 37 68
0
π ≈ . .
º
113.
14.
15.
point
a in= + ( ) =
− +( ) =
5 72
13 78 2
180 65 15º º º 1180 80 100
115
º º º
º
− =
∠ = ∠ =
16.
17.
obtuse
:
cor
m cLM m eHL
rresponding angles
m MLK m eHL∠ = − ∠ =
− =
180
180 115 65
º
º º ºº
º
supplementary angles
:
corre
18. m acL m MLK∠ = ∠ = 65
ssponding angles
:
correspondi
19. ∠ = ∠ =ceD m eHL 115º
nng angles
:
su
20. m eHG m eHL∠ = − ∠ =
− =
180
180 115 65
º
º º º
ppplementary angles
Systematic Review 14C1.
2.
V m
V Bh
= ( ) ( ) ( ) =
= =
3 5 4 60 3
ππ ≈
π ≈
r h
V Bh r h
2
3 14 42 4 200 96 3
2
3 14
. . ft
.
( ) ( )( ) =
= =
(3.
)) ( ) ( ) =
= ( ) ( ) =
=
62 10 1 130 4 3
8 8 64 2
2
, . ft
4.
5.
a cm
a rπ ≈ 33 14 62 113 04 2
3 5 3 5 3 14
. .
.
( ) ( ) =
= ( ) ( ) ( ) ( ) ( ) (cm
a6. π ≈ )) =
−( ) => ( ) −( ) =
( ) ( ) =
47 1 2
2 180 8 2 180
6 180 1
.
º º
º
cm
n7.
,, º
, º º
º º
080
1 080 8 135
180 135
total
÷ =
− =
per angle
8. 445 360 8 45
45 8 360
360 10 36
º º º;
º º
º º
or
sid
÷
÷
=
( ) =
=9. ees
c r in
10.
11.
12.
sphere
= ( ) ( ) ( ) =2 2 3 14 6 37 68
0
π ≈ . .
º
113.
14.
15.
point
a in= + ( ) =
− +( ) =
5 72
13 78 2
180 65 15º º º 1180 80 100
115
º º º
º
− =
∠ = ∠ =
16.
17.
obtuse
:
cor
m cLM m eHL
rresponding angles
m MLK m eHL∠ = − ∠ =
− =
180
180 115 65
º
º º ºº
º
supplementary angles
:
corre
18. m acL m MLK∠ = ∠ = 65
ssponding angles
:
correspondi
19. ∠ = ∠ =ceD m eHL 115º
nng angles
:
su
20. m eHG m eHL∠ = − ∠ =
− =
180
180 115 65
º
º º º
ppplementary angles
Systematic Review 14DSystematic Review 14D1.
2.
V m
V B
= ( ) ( ) ( ) =
=
7 10 8 560 3
hh r h
V Bh r h
=
( ) ( ) ( ) =
= =
π ≈
π ≈
2
3 14 3 52 3 115 40 3
2
3
. . . ft
3.
.. .
. . .
14 42 7 351 68 3
5 4 4 1 22 14
( ) ( )( ) =
= ( ) ( ) =
in
a cm4. 22
2 3 14 112 379 94 2
10 4
5.
6.
a r cm
V
= ( ) ( ) =
= ( ) ( )( )π ≈
π ≈
. .
110 4 3 14 125 6 2
2 180 10 2 18
( ) ( ) ( ) =
−( ) => ( ) −( ). .
º
cm
n7. 00
8 180 1 440
1 440 10 144
180 144 3
º
º , º
, º º
º º
=
( ) ( ) =
=
− =
÷
8. 66
360 10 36
36 10 360
360 45 8
º
º º;
º º
º º
or ÷
÷
=
× =
=9.
10. diaameter
23º
plane
yes, because
11.
12.
13.
14.
0
3 4 5
º
+ >
115.
16.
m m Bca
m
∠ = ∠ =
∠
10 62
10
º
definition of bisector
++ ∠ = + =
∠ = − ∠ + ∠( ) =
m
m m m
11 62 62 124
9 180 10 11
180
º º º
º
º
17.
−− ( ) =
∠ =
124 56
9 56
º º
º
supplementary angles
from18. m problem 17
: corresponding angles
( )∠ =m 1 56
6
º
19. 22 7º: m m Bca∠ = ∠ because they
are alternate interioor angles.
: vertical angles20. m m
m m
∠ = ∠
∠ = ∠
11 12
7 111 58
7 12 58
=
∠ = ∠ =
º:
º
alternate interior
or:
: corm m rresponding
sYsteMatic reVieW 14D - Lesson Practice 15a
soLUtions GeoMetrY158
Systematic Review 14D1.
2.
V m
V B
= ( ) ( ) ( ) =
=
7 10 8 560 3
hh r h
V Bh r h
=
( ) ( ) ( ) =
= =
π ≈
π ≈
2
3 14 3 52 3 115 40 3
2
3
. . . ft
3.
.. .
. . .
14 42 7 351 68 3
5 4 4 1 22 14
( ) ( )( ) =
= ( ) ( ) =
in
a cm4. 22
2 3 14 112 379 94 2
10 4
5.
6.
a r cm
V
= ( ) ( ) =
= ( ) ( )( )π ≈
π ≈
. .
110 4 3 14 125 6 2
2 180 10 2 18
( ) ( ) ( ) =
−( ) => ( ) −( ). .
º
cm
n7. 00
8 180 1 440
1 440 10 144
180 144 3
º
º , º
, º º
º º
=
( ) ( ) =
=
− =
÷
8. 66
360 10 36
36 10 360
360 45 8
º
º º;
º º
º º
or ÷
÷
=
× =
=9.
10. diaameter
23º
plane
yes, because
11.
12.
13.
14.
0
3 4 5
º
+ >
115.
16.
m m Bca
m
∠ = ∠ =
∠
10 62
10
º
definition of bisector
++ ∠ = + =
∠ = − ∠ + ∠( ) =
m
m m m
11 62 62 124
9 180 10 11
180
º º º
º
º
17.
−− ( ) =
∠ =
124 56
9 56
º º
º
supplementary angles
from18. m problem 17
: corresponding angles
( )∠ =m 1 56
6
º
19. 22 7º: m m Bca∠ = ∠ because they
are alternate interioor angles.
: vertical angles20. m m
m m
∠ = ∠
∠ = ∠
11 12
7 111 58
7 12 58
=
∠ = ∠ =
º:
º
alternate interior
or:
: corm m rresponding
Systematic Review 14ESystematic Review 14E1.
2
V m= ( ) ( ) ( ) =20 30 10 6 000 3,
..
3.
V Bh r h
V Bh r
= =
( ) ( ) ( ) =
= =
π ≈
π
2
3 14 102 15 4 710 3
2
. , ft
hh
in
a bh
≈
3 14 52 7 549 5 3
12
12
8 6 24
. .( ) ( )( ) =
= = ( ) ( ) =4. iin
a r cm
a
2
2 3 14 42 50 24 2
12
20 12
15
5.
6.
= ( ) ( ) =
= ( ) ×
π ≈ . .
(( ) ×
( ) ( ) ( ) =
−( ) =>
π ≈
10 7 5 3 14 235 5 2
2 180
. . .
º
units
n7. 66 2 180
4 180 720
720 6 120
180 120
( ) −( ) =
( ) =
=
−
º
º º
º º
º º
÷
8. == =60 360 6 60º º ºor ÷
360º is always the total
of eexterior degrees.
tangent
9.
10.
1
360 60 6º º÷ = sides
11.
12.
13.
c r in= ( ) ( ) ( ) =2 2 3 14 15 94 2
60
π ≈ . .
'
line or lline segment or ray( )= + + + =14.
15
P 25 25 25 25 100 ft
..
16.
180 23 35
180 58 122
23
º º º
º º º
º;
− +( ) =
− =
=minor arc
mmajor arc º º º= − =
=
=
=
360 23 337
16 4
100 10
25 5
17.
18.
19.
220. 144 12=
Systematic Review 14E1.
2
V m= ( ) ( ) ( ) =20 30 10 6 000 3,
..
3.
V Bh r h
V Bh r
= =
( ) ( ) ( ) =
= =
π ≈
π
2
3 14 102 15 4 710 3
2
. , ft
hh
in
a bh
≈
3 14 52 7 549 5 3
12
12
8 6 24
. .( ) ( )( ) =
= = ( ) ( ) =4. iin
a r cm
a
2
2 3 14 42 50 24 2
12
20 12
15
5.
6.
= ( ) ( ) =
= ( ) ×
π ≈ . .
(( ) ×
( ) ( ) ( ) =
−( ) =>
π ≈
10 7 5 3 14 235 5 2
2 180
. . .
º
units
n7. 66 2 180
4 180 720
720 6 120
180 120
( ) −( ) =
( ) =
=
−
º
º º
º º
º º
÷
8. == =60 360 6 60º º ºor ÷
360º is always the total
of eexterior degrees.
tangent
9.
10.
1
360 60 6º º÷ = sides
11.
12.
13.
c r in= ( ) ( ) ( ) =2 2 3 14 15 94 2
60
π ≈ . .
'
line or lline segment or ray( )= + + + =14.
15
P 25 25 25 25 100 ft
..
16.
180 23 35
180 58 122
23
º º º
º º º
º;
− +( ) =
− =
=minor arc
mmajor arc º º º= − =
=
=
=
360 23 337
16 4
100 10
25 5
17.
18.
19.
220. 144 12=
Lesson Practice 15ALesson Practice15A1.
2.
3.
slant height
altitude
verrtex
13circle
congruent, parallel
paralle
4.
5.
6.
7. llograms
43
8.
9.
10.
πr
V Bh in
V
3
13
13
5 5 6 50 3= = ( ) ( ) ( ) =
= 113
13
3 14 32 11 103 62 3
12
8 9
Bh
in
V Bh
≈
. .( ) ( )( ) =
= = ( )11. (( ) ( ) =
=
( ) ( ) =
15 540 3
43
3
43
3 14 23 33 49
ft
. . ft
12. V rπ ≈
33 in this
and in other problems of this
type, yyou may ignore small
answer variations caused bby
differences in rounding technique.
13. V Bh= =13
113
10 10 40
1 333 33 3
13
13
2
13
( ) ( ) ( )
= =
≈
π ≈
, . in
V Bh r h14.
33 14 22 6 25 12 3. .( ) ( ) ( ) = in
Lesson Practice 15B - sYsteMatic reVieW 15c
soLUtionsGeoMetrY 159
Lesson Practice 15BLesson Practice15B1.
2.
3.
prism
volume; sphere
basee; height
altitude
13face
4.
5.
6.
7.
8.
9.
13
1
square
V =33
13
3 6 3 6 4
17 28 3
13
13
3 14
Bh
V Bh
= ( ) ( ) ( )
=
=
(
. .
. ft
.
10. ≈
)) ( ) ( ) =
= = ( ) ( )
4 22 9 7 179 09 3
12
4 8 3 2 7 8
. . . ft
. . .11. V Bh (( )
=
= ( ) ( ) =
59 9 3
43
3 43
3 14 13 4 19 3
.
. . ft
cm
V r12.
13.
π ≈
ppyramid:
rectangul
V Bh= = ( ) ( ) ( ) =13
13
6 6 3 6 43 2 3. . ft
aar solid:
total:
V
V
= ( ) ( ) ( ) =
= +
6 6 2 7 97 2 3
43 2 97
. . ft
. .. . ft
.
2 140 4 3
13
13
2
13
3 14 6
=
= =
( ) ( )
14. cone: V Bh r hπ ≈
2214 527 52 3
2 3 14 62
( ) =
= =
( ) ( )
.
.
in
V Bh
r h
cylinder:
π ≈ 111 1 243 44 3
527 52 1 243 44
1 77
( ) =
= +
=
, .
. , .
,
in
Vtotal:
00 96 3. in
Systematic Review 15CSystematic Review 15C1. V Bh= = × × × =1
313
52
52
92
222524
9 38
3
13
13
2 5 2 5 4 5 9 375
=
= = ( )( )( ) =. . . .
m or
V Bh mm
V Bh r
i
3
13
13
2
13
3 14 4 52 8 25 174 86
2. = =
( ) ( ) ( ) =
π ≈
. . . . nn
V Bh
3
12
4 5 6 60 3
3. prism:
rectangular s
= = ( ) ( ) ( ) = ft
oolid:
total:
V
V
= ( ) ( ) ( ) =
= + =
5 6 2 60 3
60 60 120 3
ft
ft
4. VV r
in
t
= ( ) ( )
=
43
3 43
3 14 43
267 95 3
π ≈ .
.
5.
6.
altitude
heyy are parallel. (or congruent)
7. V = ( ) ( ) ( )20 20 20 ==
= ( ) ( ) =
= ( )
8 000 3
2 3 14 52 78 5 2
12
2
,
. . ft
in
a r
a
8.
9.
π ≈
×× ( ) × ( ) ( ) ( )
=
12
10 1 5 3 14
15 7 2
π ≈ .
. ft
10. circumferennce
11. n −( ) => ( ) −( ) =
( ) =
2 180 10 2 180
8 180 1 440
1
º º
º , º
,4440 10 144º º÷ =
12. if interior angle is 120º,
then exterior angles are
180º or 60º.
exterior
−120º
angles always add up
to 360º. 360 60 6º º÷ = sidess
13.
14.
use protractor to check
use protractor too check: new
angles should each measure 62.5º.
115. a radius and tangent that touch
a circle at the same point are
always perpendicular to eacch other.
m :
corresponding angles
16. ∠ = ∠ =6 14 50m º
mm m∠ = − ∠ =
− =
5 180 6
180 50 130
º
º º º:
supplementary angless
there are other valid ways of
arriving at thiis answer.
sup
17. m BcD m∠ = − ∠ =
− =
180 14
180 50 130
º
º º º:
pplementary angles
m m BcD
lin
∠ = ∠ = ( ) =10 12
12
130 65º º:
ee
m
ac bisects BcD
from last problemº
∠
∠ =18. 11 65
mm m
m m
∠ = ∠ =
∠ = − ∠
4 11 65
8 180 4
º:
º
corresponding angles
==
− =
∠ =
180 65 115
11 65
º º º:
º
supplementary angles
19. m
º:
from problem 17
vertical angles
m m∠ = ∠ =12 11 65
220. complementary; if two angles are
complementarry, they add up to 90º.
Systematic Review 15C1. V Bh= = × × × =1
313
52
52
92
222524
9 38
3
13
13
2 5 2 5 4 5 9 375
=
= = ( )( )( ) =. . . .
m or
V Bh mm
V Bh r
i
3
13
13
2
13
3 14 4 52 8 25 174 86
2. = =
( ) ( ) ( ) =
π ≈
. . . . nn
V Bh
3
12
4 5 6 60 3
3. prism:
rectangular s
= = ( ) ( ) ( ) = ft
oolid:
total:
V
V
= ( ) ( ) ( ) =
= + =
5 6 2 60 3
60 60 120 3
ft
ft
4. VV r
in
t
= ( ) ( )
=
43
3 43
3 14 43
267 95 3
π ≈ .
.
5.
6.
altitude
heyy are parallel. (or congruent)
7. V = ( ) ( ) ( )20 20 20 ==
= ( ) ( ) =
= ( )
8 000 3
2 3 14 52 78 5 2
12
2
,
. . ft
in
a r
a
8.
9.
π ≈
×× ( ) × ( ) ( ) ( )
=
12
10 1 5 3 14
15 7 2
π ≈ .
. ft
10. circumferennce
11. n −( ) => ( ) −( ) =
( ) =
2 180 10 2 180
8 180 1 440
1
º º
º , º
,4440 10 144º º÷ =
12. if interior angle is 120º,
then exterior angles are
180º or 60º.
exterior
−120º
angles always add up
to 360º. 360 60 6º º÷ = sidess
13.
14.
use protractor to check
use protractor too check: new
angles should each measure 62.5º.
115. a radius and tangent that touch
a circle at the same point are
always perpendicular to eacch other.
m :
corresponding angles
16. ∠ = ∠ =6 14 50m º
mm m∠ = − ∠ =
− =
5 180 6
180 50 130
º
º º º:
supplementary angless
there are other valid ways of
arriving at thiis answer.
sup
17. m BcD m∠ = − ∠ =
− =
180 14
180 50 130
º
º º º:
pplementary angles
m m BcD
lin
∠ = ∠ = ( ) =10 12
12
130 65º º:
ee
m
ac bisects BcD
from last problemº
∠
∠ =18. 11 65
mm m
m m
∠ = ∠ =
∠ = − ∠
4 11 65
8 180 4
º:
º
corresponding angles
==
− =
∠ =
180 65 115
11 65
º º º:
º
supplementary angles
19. m
º:
from problem 17
vertical angles
m m∠ = ∠ =12 11 65
220. complementary; if two angles are
complementarry, they add up to 90º.
sYsteMatic reVieW 15c - sYsteMatic reVieW 15D
soLUtions GeoMetrY160
Systematic Review 15C1. V Bh= = × × × =1
313
52
52
92
222524
9 38
3
13
13
2 5 2 5 4 5 9 375
=
= = ( )( )( ) =. . . .
m or
V Bh mm
V Bh r
i
3
13
13
2
13
3 14 4 52 8 25 174 86
2. = =
( ) ( ) ( ) =
π ≈
. . . . nn
V Bh
3
12
4 5 6 60 3
3. prism:
rectangular s
= = ( ) ( ) ( ) = ft
oolid:
total:
V
V
= ( ) ( ) ( ) =
= + =
5 6 2 60 3
60 60 120 3
ft
ft
4. VV r
in
t
= ( ) ( )
=
43
3 43
3 14 43
267 95 3
π ≈ .
.
5.
6.
altitude
heyy are parallel. (or congruent)
7. V = ( ) ( ) ( )20 20 20 ==
= ( ) ( ) =
= ( )
8 000 3
2 3 14 52 78 5 2
12
2
,
. . ft
in
a r
a
8.
9.
π ≈
×× ( ) × ( ) ( ) ( )
=
12
10 1 5 3 14
15 7 2
π ≈ .
. ft
10. circumferennce
11. n −( ) => ( ) −( ) =
( ) =
2 180 10 2 180
8 180 1 440
1
º º
º , º
,4440 10 144º º÷ =
12. if interior angle is 120º,
then exterior angles are
180º or 60º.
exterior
−120º
angles always add up
to 360º. 360 60 6º º÷ = sidess
13.
14.
use protractor to check
use protractor too check: new
angles should each measure 62.5º.
115. a radius and tangent that touch
a circle at the same point are
always perpendicular to eacch other.
m :
corresponding angles
16. ∠ = ∠ =6 14 50m º
mm m∠ = − ∠ =
− =
5 180 6
180 50 130
º
º º º:
supplementary angless
there are other valid ways of
arriving at thiis answer.
sup
17. m BcD m∠ = − ∠ =
− =
180 14
180 50 130
º
º º º:
pplementary angles
m m BcD
lin
∠ = ∠ = ( ) =10 12
12
130 65º º:
ee
m
ac bisects BcD
from last problemº
∠
∠ =18. 11 65
mm m
m m
∠ = ∠ =
∠ = − ∠
4 11 65
8 180 4
º:
º
corresponding angles
==
− =
∠ =
180 65 115
11 65
º º º:
º
supplementary angles
19. m
º:
from problem 17
vertical angles
m m∠ = ∠ =12 11 65
220. complementary; if two angles are
complementarry, they add up to 90º.
Systematic Review 15DSystematic Review 15D1. V Bh
x x x
= =
=
13
13
132
132
72
1 1, 88324
49 724
3
13
6 5 6 5 3 5 49 29 3
=
( )( )( )
=
. . . .
m
or
m
V
≈
2. 113
13
2
13
3 14 3 42 7 6 91 96 3
Bh r h
in
=
( )( )( ) =
π ≈
. . . .
3. conee:
cyl
V Bh r h
in
= =
( )( )( ) =
13
13
2
13
3 14 42 12 200 96 3
π ≈
. .
iinder:
total:
V Bh r h in= = ( )( )( )=π ≈2 3 14 42 10 5024 3. .
VV in
V r
= + =
= ( )(200 96 502 4 703 36 3
43
3 43
3 14 23
. . .
.4. π ≈ )) =
= = ( )
33 49 3
20 15
. in
a bh
5.
6.
7.
parallelogram
chord
(( ) =
= + + + =
= ( ) × ( )
300 2
20 20 20 20 80
12
3 12
6
in
P in
a
8.
9. ××
( )( )( ) =
−( )
π ≈
1 5 3 3 14 14 13 2
2 1
. . . ft
10.
11.
scalene
n 880 8 2 180
6 180 1 080
1080 8 135
º º
º , º
º º
=> ( ) −( ) =
( ) =
=÷
12. eexterior angles = − =
=
180 150 30
360 30 12
º º º
º º÷ sides
113.
14.
check with ruler
check with ruler:
each haalf should be 238
check with protractor
in
15.
16..
17.
18.
midpoint
aF aD x
m BFD m aF
= = =
∠ = ∠
12
12
278
1 716
ee
m BFc m BFD
=
∠ = ∠ = (
88 5
12
12
88 5
. º:
. º
vertical angles
)) =
∠ = − ∠ =
−
44 25
180
180 88 5
. º:
º
º .
bisector
19. m eFD m aFe
ºº . º:= 91 5
supplementary angles
supplementary;20. if two angles add
up to 180º, they are supplemeentary.
Systematic Review 15D1. V Bh
x x x
= =
=
13
13
132
132
72
1 1, 88324
49 724
3
13
6 5 6 5 3 5 49 29 3
=
( )( )( )
=
. . . .
m
or
m
V
≈
2. 113
13
2
13
3 14 3 42 7 6 91 96 3
Bh r h
in
=
( )( )( ) =
π ≈
. . . .
3. conee:
cyl
V Bh r h
in
= =
( )( )( ) =
13
13
2
13
3 14 42 12 200 96 3
π ≈
. .
iinder:
total:
V Bh r h in= = ( )( )( )=π ≈2 3 14 42 10 5024 3. .
VV in
V r
= + =
= ( )(200 96 502 4 703 36 3
43
3 43
3 14 23
. . .
.4. π ≈ )) =
= = ( )
33 49 3
20 15
. in
a bh
5.
6.
7.
parallelogram
chord
(( ) =
= + + + =
= ( ) × ( )
300 2
20 20 20 20 80
12
3 12
6
in
P in
a
8.
9. ××
( )( )( ) =
−( )
π ≈
1 5 3 3 14 14 13 2
2 1
. . . ft
10.
11.
scalene
n 880 8 2 180
6 180 1 080
1080 8 135
º º
º , º
º º
=> ( ) −( ) =
( ) =
=÷
12. eexterior angles = − =
=
180 150 30
360 30 12
º º º
º º÷ sides
113.
14.
check with ruler
check with ruler:
each haalf should be 238
check with protractor
in
15.
16..
17.
18.
midpoint
aF aD x
m BFD m aF
= = =
∠ = ∠
12
12
278
1 716
ee
m BFc m BFD
=
∠ = ∠ = (
88 5
12
12
88 5
. º:
. º
vertical angles
)) =
∠ = − ∠ =
−
44 25
180
180 88 5
. º:
º
º .
bisector
19. m eFD m aFe
ºº . º:= 91 5
supplementary angles
supplementary;20. if two angles add
up to 180º, they are supplemeentary.
sYsteMatic reVieW 15e - Lesson Practice 16B
soLUtionsGeoMetrY 161
Systematic Review 15ESystematic Review 15E1. V Bh= = ( ) ( )( )
=
12
4 5 4 2 11
1
. .
003 95 3
43
3 43
3 14 1 53
14 13 3
.
. .
.
mm
V r
cm
2.
3.
= ( ) ( )
=
π ≈
pyyramid:
rectangular s
V Bh= = ( ) ( ) ( ) =13
13
8 8 6 128 3ft
oolid:
total:
V
V
= ( ) ( ) ( ) =
= + =
8 8 3 192 3
128 192 320
ft
ft33
8 3 5 28 2
4.
5.
6.
7.
8.
altitude
prism
13
a bh in= = ( ) ( ) =.
PP in
a
= + + + =
= ( ) × ( ) × ( ) ( ) ( )8 4 8 4 24
12
8 12
10 4 5 3 14.9. π ≈
. ft
º
=
−( ) => ( ) −(
62 8 2
2 180 5 2
10.
11.
equilateral
n )) =
( ) =
=
180
3 180 540
540 5 108
8
º
º º
º º÷
12.
13. exterior anngles
check w
=
− =
=
180 135 45
360 45 8
º º º;
º º÷ sides
14. iith ruler
check with protractor:
angles shou
15.
lld measure 55º
check with protractor
83
16.
17. ⋅84 == + =
= − =
− = + + −
83 4 87
28 23 28 3 25
2 3 5 1 2 3 5 1
18.
19.
÷
X X Y Y X Y(( )
=
=−
= −
X Y5 4
1
10310 3
110 320.
Lesson Practice 16ALesson Practice16A1.
2.
3.
4.
6
5
4
circles; rectanglee
square
height; circumference
5.
6.
7. sa = ( ) ( ) +2 3 5 2 33 4 2 4 5
30 24 40 94 2
2 2 2
2
( ) ( ) + ( ) ( ) =
+ + =
= +
ft
8. sa r rhπ π ≈
(( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+ =
3 14 102 2 3 14 10 5
628 314 942 2
. .
ft
99.
10
sa
in
= ( ) ( ) + ( ) ( ) ( ) =
+ =
5 5 4 12
5 6
25 60 85 2
( )
.. base:
sides:
sa
sa
= ( ) ( ) =
= ( ) ( ) ( ) =
12
6 5 15 2
3 12
6 8 7
ft
22 2
15 72 87 2
2 5 4 2 5 7
ft
ft
total:
sa
sa
= + =
= ( ) ( ) + ( ) (11. )) + ( )( ) =
+ + =
= ( )
2 4 7
40 70 56 166 2
2 4 5 6
ft
"
.
12. roof":
sa (( ) =
= ( ) ( ) + ( ) ( )
= + =
54 2
2 4 5 2 4 6
40 48 88
ft
ft
sides:
sa
22
2 12
4 5 20 2
triangles:
bottom:
sa
sa
= ( ) ( ) =
=
ft
55 6 30 2
54 88 20 30 192 2
( ) ( ) =
= + + + =
ft
ft
total:
sa
Lesson Practice 16BLesson Practice16B1.
2.
3.
4.
square
cube
tri
pyramid
aangular
cylinder
rectangular
5.
6.
7. sa =
( ) ( ) +2 10 30 2 110 20 2 20 30
600 400 1 200 2 200 2
( ) ( ) + ( ) ( ) =
+ + =, , ft
8. sa == +
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
2 2 2
2 3 14 152 2 3 14 15 60
1
π π ≈r rh
. .
,4413 5 652 7 065 2
3 6 3 6 4 12
3 6
+ =
= ( ) ( ) + ( ) (, ,
. . ( ) .
in
sa9. )) ( ) =
+ =
=
( ) ( )
5 2
12 96 37 44 50 4 2
2 12
4 8
.
. . . ft
( ) .
10. sa
33 2 2 4 7 8 4 8 7 8
15 36 62 4 37 44
. . . .
. . .
( ) + ( )( ) ( ) + ( ) ( ) =
+ + = 1115 2 2
2 12
8 9 11 15 8 15
.
( )
cm
sa11. =
( ) ( ) ( ) + ( ) ( ) + ( ) ( ) ++ ( ) ( ) =
+ + + =
= (
9 15
72 165 120 135 492 2
2
" " :
in
roof
sa
12.
)) ( ) ( ) =
= ( ) ( ) ( ) =
6 6 72 2
2 12
5 4 20
ft
( ) ft
triangles:
sa 22
2 5 5 2 6 5 50 60 110 2
sides
sa
:
ft= ( ) ( ) ( ) + ( ) ( ) ( ) = + =
botttom:
total:
sa
sa
= ( ) ( ) =
= + + + =
5 6 30 2
72 20 110 30 232
ft
ft2
Lesson Practice 16B - sYsteMatic reVieW 16c
soLUtions GeoMetrY162
Lesson Practice16B1.
2.
3.
4.
square
cube
tri
pyramid
aangular
cylinder
rectangular
5.
6.
7. sa =
( ) ( ) +2 10 30 2 110 20 2 20 30
600 400 1 200 2 200 2
( ) ( ) + ( ) ( ) =
+ + =, , ft
8. sa == +
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
2 2 2
2 3 14 152 2 3 14 15 60
1
π π ≈r rh
. .
,4413 5 652 7 065 2
3 6 3 6 4 12
3 6
+ =
= ( ) ( ) + ( ) (, ,
. . ( ) .
in
sa9. )) ( ) =
+ =
=
( ) ( )
5 2
12 96 37 44 50 4 2
2 12
4 8
.
. . . ft
( ) .
10. sa
33 2 2 4 7 8 4 8 7 8
15 36 62 4 37 44
. . . .
. . .
( ) + ( )( ) ( ) + ( ) ( ) =
+ + = 1115 2 2
2 12
8 9 11 15 8 15
.
( )
cm
sa11. =
( ) ( ) ( ) + ( ) ( ) + ( ) ( ) ++ ( ) ( ) =
+ + + =
= (
9 15
72 165 120 135 492 2
2
" " :
in
roof
sa
12.
)) ( ) ( ) =
= ( ) ( ) ( ) =
6 6 72 2
2 12
5 4 20
ft
( ) ft
triangles:
sa 22
2 5 5 2 6 5 50 60 110 2
sides
sa
:
ft= ( ) ( ) ( ) + ( ) ( ) ( ) = + =
botttom:
total:
sa
sa
= ( ) ( ) =
= + + + =
5 6 30 2
72 20 110 30 232
ft
ft2
Systematic Review 16CSystematic Review 16C1. sa = ( )( ) + ( ) ( )( ) =2 2 4 1
22 4( ) 44 16
20 2
2 2 2
2 3 14 52 2 3 14
+
=
= +
( ) ( ) ( ) + ( ). .
m
sa r rh2. π π ≈
(( ) ( ) ( ) =
+ =
= ( ) (
5 12
157 376 8 533 8 2
9
. . cm
sa
3. "roof":
2 )) ( ) =
= ( ) ( ) ( ) =
12 216 2
2 12
9 8 72 2( )
m
sa m
triangles:
siddes:
bott
sa
m
= ( ) ( ) ( ) + ( ) ( ) ( )
= + =
2 6 12 2 6 9
144 108 252 2
oom:
total:
sa m
sa
= ( ) ( ) =
= + + + =
9 12 108 2
216 72 252 108 6488 2
43
3 43
3 14 1 663
19
. .
m
V r
4.
5.
slant height
= ( ) ( )π ≈
≈ ..
.
15 3
2 2 3
in
c r
6.
7.
8.
latitude
straight angle
= ( )π ≈ 114 7 43 96
2 3 14 52 12
942
( ) ( ) =
= = ( ) ( )( )
=
. ft
.9. V Bh r hπ ≈
º º
cm
n
3
2 180 12 2 180
10 18
10.
11.
secant
−( ) => ( ) −( ) =
( ) 00 1 800
1 800 12 150
180
º , º
, º º
º
=
=
= −
÷
12. exterior angle 1108 72
360 72 5
º º
º º
=
=÷ sides
check with protract13. oor
check with protractor
check with ruler
14.
15. aand protractor:
all angles should measure 90º
16..
17.
rays
aF FD and
eF, Fe, FB, BF, eB, or Be
, , aaD
m BFc m BFD18. ∠ = ∠( ) =
( ) =
12
12
90 45º º:
definition of a bisector
m∠ = ∠ + ∠ =
+ =
⊥
aFc m aFB m BFc
90 45 135º º º
,19. or is perpendicular to
congruent; if altern20. aate interior
angles are congruent, they are
forrmed by parallel
lines cut by a transversal.
Systematic Review 16C1. sa = ( )( ) + ( ) ( )( ) =2 2 4 1
22 4( ) 44 16
20 2
2 2 2
2 3 14 52 2 3 14
+
=
= +
( ) ( ) ( ) + ( ). .
m
sa r rh2. π π ≈
(( ) ( ) ( ) =
+ =
= ( ) (
5 12
157 376 8 533 8 2
9
. . cm
sa
3. "roof":
2 )) ( ) =
= ( ) ( ) ( ) =
12 216 2
2 12
9 8 72 2( )
m
sa m
triangles:
siddes:
bott
sa
m
= ( ) ( ) ( ) + ( ) ( ) ( )
= + =
2 6 12 2 6 9
144 108 252 2
oom:
total:
sa m
sa
= ( ) ( ) =
= + + + =
9 12 108 2
216 72 252 108 6488 2
43
3 43
3 14 1 663
19
. .
m
V r
4.
5.
slant height
= ( ) ( )π ≈
≈ ..
.
15 3
2 2 3
in
c r
6.
7.
8.
latitude
straight angle
= ( )π ≈ 114 7 43 96
2 3 14 52 12
942
( ) ( ) =
= = ( ) ( )( )
=
. ft
.9. V Bh r hπ ≈
º º
cm
n
3
2 180 12 2 180
10 18
10.
11.
secant
−( ) => ( ) −( ) =
( ) 00 1 800
1 800 12 150
180
º , º
, º º
º
=
=
= −
÷
12. exterior angle 1108 72
360 72 5
º º
º º
=
=÷ sides
check with protract13. oor
check with protractor
check with ruler
14.
15. aand protractor:
all angles should measure 90º
16..
17.
rays
aF FD and
eF, Fe, FB, BF, eB, or Be
, , aaD
m BFc m BFD18. ∠ = ∠( ) =
( ) =
12
12
90 45º º:
definition of a bisector
m∠ = ∠ + ∠ =
+ =
⊥
aFc m aFB m BFc
90 45 135º º º
,19. or is perpendicular to
congruent; if altern20. aate interior
angles are congruent, they are
forrmed by parallel
lines cut by a transversal.
sYsteMatic reVieW 16c - sYsteMatic reVieW 16e
soLUtionsGeoMetrY 163
Systematic Review 16C1. sa = ( )( ) + ( ) ( )( ) =2 2 4 1
22 4( ) 44 16
20 2
2 2 2
2 3 14 52 2 3 14
+
=
= +
( ) ( ) ( ) + ( ). .
m
sa r rh2. π π ≈
(( ) ( ) ( ) =
+ =
= ( ) (
5 12
157 376 8 533 8 2
9
. . cm
sa
3. "roof":
2 )) ( ) =
= ( ) ( ) ( ) =
12 216 2
2 12
9 8 72 2( )
m
sa m
triangles:
siddes:
bott
sa
m
= ( ) ( ) ( ) + ( ) ( ) ( )
= + =
2 6 12 2 6 9
144 108 252 2
oom:
total:
sa m
sa
= ( ) ( ) =
= + + + =
9 12 108 2
216 72 252 108 6488 2
43
3 43
3 14 1 663
19
. .
m
V r
4.
5.
slant height
= ( ) ( )π ≈
≈ ..
.
15 3
2 2 3
in
c r
6.
7.
8.
latitude
straight angle
= ( )π ≈ 114 7 43 96
2 3 14 52 12
942
( ) ( ) =
= = ( ) ( )( )
=
. ft
.9. V Bh r hπ ≈
º º
cm
n
3
2 180 12 2 180
10 18
10.
11.
secant
−( ) => ( ) −( ) =
( ) 00 1 800
1 800 12 150
180
º , º
, º º
º
=
=
= −
÷
12. exterior angle 1108 72
360 72 5
º º
º º
=
=÷ sides
check with protract13. oor
check with protractor
check with ruler
14.
15. aand protractor:
all angles should measure 90º
16..
17.
rays
aF FD and
eF, Fe, FB, BF, eB, or Be
, , aaD
m BFc m BFD18. ∠ = ∠( ) =
( ) =
12
12
90 45º º:
definition of a bisector
m∠ = ∠ + ∠ =
+ =
⊥
aFc m aFB m BFc
90 45 135º º º
,19. or is perpendicular to
congruent; if altern20. aate interior
angles are congruent, they are
forrmed by parallel
lines cut by a transversal.
Systematic Review 16DSystematic Review 16D1. sa =
( ) ( ) ( ) + ( )2 12
9 4 2 2 6( ) . .22 11 9 11
37 8 136 4 99 273 2 2
2
( ) ( ) + ( ) ( ) =
+ + =
=
. . . mm
sa2. πrr rh2 2
2 3 14 82 2 3 14 8 10
401 92
+
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+
π ≈
. .
. 5502 4 904 32 2
4 12
6 5 2
. .
( ) .
=
= ( ) ( ) ( )
cm
sa
3. triangles:
==
= ( ) ( ) ( ) =
62 4 2
4 6 2 7 64 8 2
. ft
. . ft
sides:
bottom:
sa
saa = ( ) ( ) =
+ + =
6 6 36 2
62 4 64 8 36 163 2 2
ft
. . . ft
total:
a4. lltitude
5.
6.
V r
in
V
= ( ) ( )
=
=
43
3 43
3 14 1 53
14 13 3
π ≈ . .
.
113
13
2
13
3 14 2 82 4 7 38 57 3
Bh r h
cm
=
( ) ( )( )
π ≈
≈. . . .
7. refllex angle
48.
9. a = ( ) × ( ) ×
( ) ( ) ( ) =
12
10 12
20
5 10 3 14
π ≈
. 1157 2
2 180 5 2 180
3 1
º º
in
n
10.
11.
sector
−( ) => ( ) −( ) =
( ) 880 540
2 180 360
180 360 360180
º º
º º
º º º
=
−( ) ( ) =
− =
12. n
nnºº º
ºº
=
= =
720720180
4n sides
13.
14.
check with ruler
ccheck with protractor:
Fourth angle is 60º.
15. iinterior angles of a quadrilateral
add up to 3660º, so fourth angle
has a measure of:
360 65º º− ++ +( ) =
− =
⊥
115 120
360 300 60
90
º º
º º º
º16. , because aD ecc→ →
17.
18.
complementary; they add up to 90
supp
º
llementary; they form
a straight line
19. m BGc
m
∠ =
∠∠ − ∠ + ∠( ) =
− +( ) =
− =
eGc m eGF m FGB
180 34 90
180 124 56
º º º
º º ºº
20. perpendicular; if 90º angles
are formed froom intersecting
lines, the lines are perpendicuular.
Systematic Review 16D1. sa =
( ) ( ) ( ) + ( )2 12
9 4 2 2 6( ) . .22 11 9 11
37 8 136 4 99 273 2 2
2
( ) ( ) + ( ) ( ) =
+ + =
=
. . . mm
sa2. πrr rh2 2
2 3 14 82 2 3 14 8 10
401 92
+
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+
π ≈
. .
. 5502 4 904 32 2
4 12
6 5 2
. .
( ) .
=
= ( ) ( ) ( )
cm
sa
3. triangles:
==
= ( ) ( ) ( ) =
62 4 2
4 6 2 7 64 8 2
. ft
. . ft
sides:
bottom:
sa
saa = ( ) ( ) =
+ + =
6 6 36 2
62 4 64 8 36 163 2 2
ft
. . . ft
total:
a4. lltitude
5.
6.
V r
in
V
= ( ) ( )
=
=
43
3 43
3 14 1 53
14 13 3
π ≈ . .
.
113
13
2
13
3 14 2 82 4 7 38 57 3
Bh r h
cm
=
( ) ( )( )
π ≈
≈. . . .
7. refllex angle
48.
9. a = ( ) × ( ) ×
( ) ( ) ( ) =
12
10 12
20
5 10 3 14
π ≈
. 1157 2
2 180 5 2 180
3 1
º º
in
n
10.
11.
sector
−( ) => ( ) −( ) =
( ) 880 540
2 180 360
180 360 360180
º º
º º
º º º
=
−( ) ( ) =
− =
12. n
nnºº º
ºº
=
= =
720720180
4n sides
13.
14.
check with ruler
ccheck with protractor:
Fourth angle is 60º.
15. iinterior angles of a quadrilateral
add up to 3660º, so fourth angle
has a measure of:
360 65º º− ++ +( ) =
− =
⊥
115 120
360 300 60
90
º º
º º º
º16. , because aD ecc→ →
17.
18.
complementary; they add up to 90
supp
º
llementary; they form
a straight line
19. m BGc
m
∠ =
∠∠ − ∠ + ∠( ) =
− +( ) =
− =
eGc m eGF m FGB
180 34 90
180 124 56
º º º
º º ºº
20. perpendicular; if 90º angles
are formed froom intersecting
lines, the lines are perpendicuular.
Systematic Review 16ESystematic Review 16E1. sa =
( ) ( ) ( ) + ( ) ( ) ( )2 5 15 2 5 25 ++ ( ) ( ) ( ) =
+ + =
= +
2 15 25
150 250 750 1 150 2
2 2 2
, in
sa r2. ≠ πrrh ≈
2 3 14 42 10 3 14 8
100 48 251 2 35
( ) ( ) ( ) + ( ) ( ) ( ) =
+ =
. .
. . 11 68 2
4 12
8 9 144 2
.
( ) ft
cm
sa
3. "roof":
sides
= ( ) ( ) ( ) =
::
bottom:
tot
sa
sa
= ( ) ( ) ( ) =
= ( ) ( ) =
4 8 3 96 2
8 8 64 2
ft
ft
aal:
144 96 64 304 2+ + = ft
4.
5.
6.
7.
8.
9.
10.
11.
c
d
a
e
h
b
j
f
112.
13.
14.
15
i
g
m aoB
measure of minor arc aB =
∠ = 40º
.. measure of minor arc aBc =
∠ + ∠ =
+
m aoB m Boc
40 100º º ==
=
∠ + ∠ + ∠ =
140º
16. measure of arc aBcD
m aoB m Boc m coD
440 100 40 180
4
8
8 322 4
º º º º+ + =
+
× +
+
+
17. X
X
X
X X
X X
XX
X
X X
2 12 32
52
2 102 5
+ +
+× +
+
+
18.
X X
XX
X
X X
X X
2 7 10
31
32 3
2 2 3
+ +
+× −
− −
+
+ −
19.
20. XX
X
X
X X
X X
−
× +
−
−
+ −
4
6
6 242 4
2 2 24
sYsteMatic reVieW 16e - Lesson Practice 17a
soLUtions GeoMetrY164
Systematic Review 16E1. sa =
( ) ( ) ( ) + ( ) ( ) ( )2 5 15 2 5 25 ++ ( ) ( ) ( ) =
+ + =
= +
2 15 25
150 250 750 1 150 2
2 2 2
, in
sa r2. ≠ πrrh ≈
2 3 14 42 10 3 14 8
100 48 251 2 35
( ) ( ) ( ) + ( ) ( ) ( ) =
+ =
. .
. . 11 68 2
4 12
8 9 144 2
.
( ) ft
cm
sa
3. "roof":
sides
= ( ) ( ) ( ) =
::
bottom:
tot
sa
sa
= ( ) ( ) ( ) =
= ( ) ( ) =
4 8 3 96 2
8 8 64 2
ft
ft
aal:
144 96 64 304 2+ + = ft
4.
5.
6.
7.
8.
9.
10.
11.
c
d
a
e
h
b
j
f
112.
13.
14.
15
i
g
m aoB
measure of minor arc aB =
∠ = 40º
.. measure of minor arc aBc =
∠ + ∠ =
+
m aoB m Boc
40 100º º ==
=
∠ + ∠ + ∠ =
140º
16. measure of arc aBcD
m aoB m Boc m coD
440 100 40 180
4
8
8 322 4
º º º º+ + =
+
× +
+
+
17. X
X
X
X X
X X
XX
X
X X
2 12 32
52
2 102 5
+ +
+× +
+
+
18.
X X
XX
X
X X
X X
2 7 10
31
32 3
2 2 3
+ +
+× −
− −
+
+ −
19.
20. XX
X
X
X X
X X
−
× +
−
−
+ −
4
6
6 242 4
2 2 24
Systematic Review 16E1. sa =
( ) ( ) ( ) + ( ) ( ) ( )2 5 15 2 5 25 ++ ( ) ( ) ( ) =
+ + =
= +
2 15 25
150 250 750 1 150 2
2 2 2
, in
sa r2. ≠ πrrh ≈
2 3 14 42 10 3 14 8
100 48 251 2 35
( ) ( ) ( ) + ( ) ( ) ( ) =
+ =
. .
. . 11 68 2
4 12
8 9 144 2
.
( ) ft
cm
sa
3. "roof":
sides
= ( ) ( ) ( ) =
::
bottom:
tot
sa
sa
= ( ) ( ) ( ) =
= ( ) ( ) =
4 8 3 96 2
8 8 64 2
ft
ft
aal:
144 96 64 304 2+ + = ft
4.
5.
6.
7.
8.
9.
10.
11.
c
d
a
e
h
b
j
f
112.
13.
14.
15
i
g
m aoB
measure of minor arc aB =
∠ = 40º
.. measure of minor arc aBc =
∠ + ∠ =
+
m aoB m Boc
40 100º º ==
=
∠ + ∠ + ∠ =
140º
16. measure of arc aBcD
m aoB m Boc m coD
440 100 40 180
4
8
8 322 4
º º º º+ + =
+
× +
+
+
17. X
X
X
X X
X X
XX
X
X X
2 12 32
52
2 102 5
+ +
+× +
+
+
18.
X X
XX
X
X X
X X
2 7 10
31
32 3
2 2 3
+ +
+× −
− −
+
+ −
19.
20. XX
X
X
X X
X X
−
× +
−
−
+ −
4
6
6 242 4
2 2 24
Lesson Practice 17ALesson Practice17A1. 5 2 3 5 5 2 3 5+ = + :
cannot be simpplified
cannot be simplified2.
3.
8 7 3 3
12 6 10
+
− 66 12 10 6 2 6
11 2 3 2 5 2
11 3 5 2 19 2
12 24
6 3
= −( ) =
+ + =
+ +( ) =
4.
5. == = =
= ( ) =
= =
= =
12 86
2 8
2 4 2 2 2 2 4 2
25 10
5 5
25 25
5 2
24 4 6 2
6.
7. 66
300 100 3 10 3
48 16 3 4 3
5 3 6 5 5 6 3 5
8.
9.
10.
= =
= =
( )( ) = ( ) ( ) ==
( )( ) = ( ) ( ) =
= =
( ) =
30 15
6 6 7 2 6 7 6 2
42 12 42 4 3
42 2 3 84 3
11.
112.
13.
14.
2 3 2 3 2 2 3 3
4 9 4 3 12
5 2 24
3 1
( )( ) = ( )( ) =
= ( ) =
≈
≈
.
..
.
73
14 3 7415.
16.
17.
18.
≈
radical
numbers; radicals
ssquare
Lesson Practice17A1. 5 2 3 5 5 2 3 5+ = + :
cannot be simpplified
cannot be simplified2.
3.
8 7 3 3
12 6 10
+
− 66 12 10 6 2 6
11 2 3 2 5 2
11 3 5 2 19 2
12 24
6 3
= −( ) =
+ + =
+ +( ) =
4.
5. == = =
= ( ) =
= =
= =
12 86
2 8
2 4 2 2 2 2 4 2
25 10
5 5
25 25
5 2
24 4 6 2
6.
7. 66
300 100 3 10 3
48 16 3 4 3
5 3 6 5 5 6 3 5
8.
9.
10.
= =
= =
( )( ) = ( ) ( ) ==
( )( ) = ( ) ( ) =
= =
( ) =
30 15
6 6 7 2 6 7 6 2
42 12 42 4 3
42 2 3 84 3
11.
112.
13.
14.
2 3 2 3 2 2 3 3
4 9 4 3 12
5 2 24
3 1
( )( ) = ( )( ) =
= ( ) =
≈
≈
.
..
.
73
14 3 7415.
16.
17.
18.
≈
radical
numbers; radicals
ssquare
Lesson Practice 17B - sYsteMatic reVieW 17c
soLUtionsGeoMetrY 165
Lesson Practice 17BLesson Practice17B1. 6 7 5 3 6 7 5 3+ = + :
cannot be simpplified
2.
3.
4.
8 3 5 3 8 5 3 13 3
8 7 7 7 8 7 7 1 7 7
+ = +( ) =
− = −( ) = =
113 2 11 2 20 2
13 11 20 2 4 2
36 8
6 2
36 46
6 4 6 2
+ − =
+ −( ) =
= = = ( )5. ==
= =
= =
12
42 10
7 5
42 27
6 2
108 36 3 6 3
6.
7.
this can also bee done
in smaller steps:
108 = = =
( ) ( ) =
9 12 9 4 3
3 2 3 6 3
88.
9.
10.
250 25 10 5 10
180 36 5 6 5
6 7 5 7 6 5
= =
= =
( ) −( ) = ( ) −( ) 77 7
30 49 30 7 210
6 2 4 3 6 4 2 3 24 6
=
− = −( ) ( ) = −
( )( ) = ( ) ( ) =11.
112.
13.
8 3 20 8 3 20 24 20
24 4 5 24 2 5 48 5
6
( ) ( ) = ( ) ( ) = =
= ( ) =
≈ 22 45
11 3 32
21 4 58
3 3 9 3
.
.
.
14.
15.
16.
17.
18.
≈
≈
same
who
= =
lle
Systematic Review 17C Systematic Review 17C1.
2.
4 3 5 3 4 5 3 9 3
7 3 7 1 7
+ = +( ) =
+ = ++ =
+( ) =
− = −( ) =
( )( ) =
3 7
1 3 7 4 7
16 2 8 2 16 8 2 8 2
7 10 70
3.
4.
5. 55 2 3 2 5 3 2 15
28
7
41
4 2
8 4 2 2 2
28 4 7 2 7
( )( ) = =
= = =
= =
= =
6.
7.
8.
99.
10.
2 2 2 83 2 7 5 29≈ ≈. ; .
top:
V rea of base triang= a lle height
bottom:
× =
( ) ( ) × =
= ( ) ( )
12
9 8 15 540 3
9 15
m
V 77 945 3
540 945 1 485 3
( ) =
= + = ,
m
V m
sa
total:
"roof":11.
== ( ) ( ) ( ) =
= ( ) ( )
2 10 15 300 2
2 12
9
m
sa
triangles:
88 72 2
2 7 15 2 7 9
210 126
( ) =
= ( )( ) ( ) + ( )( ) ( ) =
+
m
sa
sides:
==
= ( ) ( ) =
= +
336 2
9 15 135 2
300 72
m
sa m
sa
bottom:
total:
++ + =
= =
( ) ( )
336 135 843 2
13
13
2
13
3 14 112 1.
m
V Bh r h12. π ≈
44 3 1 811 05 3
43
3 43
3 14 2 43
. , . ft
. .
( )
= ( ) ( )≈
π ≈
≈
13. V r
557 88 3
6 6 36 2
2 3 14 3
.
.
in
a bh in
a r
14.
15.
= = ( ) ( ) =
= ( )π ≈ 22 28 26 2
36 28 26 7 74 2
( ) =
= − =
.
. .
in
a in
tMs
16.
17.
18. veertical
alternate interior
70º; vertical a
19.
20. nngles
Systematic Review 17C1.
2.
4 3 5 3 4 5 3 9 3
7 3 7 1 7
+ = +( ) =
+ = ++ =
+( ) =
− = −( ) =
( )( ) =
3 7
1 3 7 4 7
16 2 8 2 16 8 2 8 2
7 10 70
3.
4.
5. 55 2 3 2 5 3 2 15
28
7
41
4 2
8 4 2 2 2
28 4 7 2 7
( )( ) = =
= = =
= =
= =
6.
7.
8.
99.
10.
2 2 2 83 2 7 5 29≈ ≈. ; .
top:
V rea of base triang= a lle height
bottom:
× =
( ) ( ) × =
= ( ) ( )
12
9 8 15 540 3
9 15
m
V 77 945 3
540 945 1 485 3
( ) =
= + = ,
m
V m
sa
total:
"roof":11.
== ( ) ( ) ( ) =
= ( ) ( )
2 10 15 300 2
2 12
9
m
sa
triangles:
88 72 2
2 7 15 2 7 9
210 126
( ) =
= ( )( ) ( ) + ( )( ) ( ) =
+
m
sa
sides:
==
= ( ) ( ) =
= +
336 2
9 15 135 2
300 72
m
sa m
sa
bottom:
total:
++ + =
= =
( ) ( )
336 135 843 2
13
13
2
13
3 14 112 1.
m
V Bh r h12. π ≈
44 3 1 811 05 3
43
3 43
3 14 2 43
. , . ft
. .
( )
= ( ) ( )≈
π ≈
≈
13. V r
557 88 3
6 6 36 2
2 3 14 3
.
.
in
a bh in
a r
14.
15.
= = ( ) ( ) =
= ( )π ≈ 22 28 26 2
36 28 26 7 74 2
( ) =
= − =
.
. .
in
a in
tMs
16.
17.
18. veertical
alternate interior
70º; vertical a
19.
20. nngles
sYsteMatic reVieW 17D - Lesson Practice 18a
soLUtions GeoMetrY166
Systematic Review 17DSystematic Review 17D1.
2.
9 6 23 6 9 23 6 32 6
7 2 8
+ = +( ) =
+ 22 7 8 2 15 2
7 5 7 1 7 5 7
1 5 7 4 7
3 6 2 7
= +( ) =
− = − =
−( ) = −
( )( ) =
3.
4. 33 2 6 7 6 42
11 11 121 11
2 30
5
2 61
2 6
12
( ) ( ) =
( )( ) = =
= =
5.
6.
7. == =
= =
=
4 3 2 3
200 100 2 10 2
2 3 3 46 10 2 14 14
8.
9.
10.
≈ ≈. ; .
V Bhh r h
cm
sa r r
= ( ) ( ) ( )
=
= +
π ≈
π π
2 3 14 22 16
200 96 3
2 2 2
.
.
11. hh ≈
2 3 14 22 2 3 14 2 16
25 12 200 96
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+ =
. .
. . 2226 08 2
4 52
6 27 2
2 3 14
.
.
cm
a units
a r
12.
13.
= + ( ) =
= (π ≈ )) ( ) ⊕
= −
2 42 18 09 2
180
. .
º
in
arc14.
15. interior angle 445 135
540 2 180
540 180 360
900 1
º º
º º
º º º
º
=
= −( )= −
=
16. n
n
880
900180
5
n
n sides
º
ºº
= =
17. cFD or BFc or aFD: alll are 90º
5 cm
: line BD is a transvers
18.
19. aDF aal cutting
lines aD and Bc, which are parallel,,
because they are opposite sides
of a rhombus.. cBF and aDF
are alternate interior angles.
∠ ∠
220. 30º; alternate interior angles
Systematic Review 17ESystematic Review 17E1. 5 5 2 2 5 5 2 2+ = + :
cannot be siimplified
3 52.
3.
+ = +( ) =
− =
−( )
8 5 3 8 5 11 5
13 10 15 10
13 15 110 2 10
5 10 3 13 5 3 10 13
15 130
3 8 2 8
= −
( )( ) = ( ) ( ) =
( )( )
4.
5. == ( ) ( ) =
= ( ) =
= = =
=
3 2 8 8
6 64 6 8 48
8 12
2 6
8 22
4 21
4 2
27 9 3
6.
7. ==
= =
= = (
3 3
20 4 5 2 5
3 3 5 20 2 5 4 47
13
13
21
8.
9.
10.
≈ ≈. ; .
V Bh ))( ) ( ) =
= ( ) ( )
21 13
1 911 3
4 12
21
,
( )
mm
sa
11. triangles:
115 630
21 21 441
630 441 1
( ) =
= ( )( ) =
= + =
base:
total:
sa
sa ,,
.
071 2
11 8 88 2
3 14
mm
a bh units
c d
12.
13.
= = ( ) ( ) =
= (π ≈ )) ( )
=
6 4 20 10
1
. .≈ in
14.
15.
isosceles
interior angle 880 72 108
360 72 5
2 10 16
º º º
º º
− =
=
+ + = +
16.
17.
÷ sides
X X X 88 2
2 5 6 2 3
2 8 7 7
( ) +( )
+ + = +( ) +( )
+ + = +( )
X
X X X X
X X X X
18.
19. ++( )
+ + = +( ) +( )1
2 9 20 5 420. X X X X
Systematic Review 17E1. 5 5 2 2 5 5 2 2+ = + :
cannot be siimplified
3 52.
3.
+ = +( ) =
− =
−( )
8 5 3 8 5 11 5
13 10 15 10
13 15 110 2 10
5 10 3 13 5 3 10 13
15 130
3 8 2 8
= −
( )( ) = ( ) ( ) =
( )( )
4.
5. == ( ) ( ) =
= ( ) =
= = =
=
3 2 8 8
6 64 6 8 48
8 12
2 6
8 22
4 21
4 2
27 9 3
6.
7. ==
= =
= = (
3 3
20 4 5 2 5
3 3 5 20 2 5 4 47
13
13
21
8.
9.
10.
≈ ≈. ; .
V Bh ))( ) ( ) =
= ( ) ( )
21 13
1 911 3
4 12
21
,
( )
mm
sa
11. triangles:
115 630
21 21 441
630 441 1
( ) =
= ( )( ) =
= + =
base:
total:
sa
sa ,,
.
071 2
11 8 88 2
3 14
mm
a bh units
c d
12.
13.
= = ( ) ( ) =
= (π ≈ )) ( )
=
6 4 20 10
1
. .≈ in
14.
15.
isosceles
interior angle 880 72 108
360 72 5
2 10 16
º º º
º º
− =
=
+ + = +
16.
17.
÷ sides
X X X 88 2
2 5 6 2 3
2 8 7 7
( ) +( )
+ + = +( ) +( )
+ + = +( )
X
X X X X
X X X X
18.
19. ++( )
+ + = +( ) +( )1
2 9 20 5 420. X X X X
Lesson Practice 18A Lesson Practice18A1.
2.
3.
4.
right
legs
hypotenuse
Pyythagorean
the hypotenuse squared;
if the leg
5.
squared plus the leg
squared equals the hypoteenuse
squared, the triangle is a
right triangle..
6.
7.
82 92 2
64 81 2
145 2
145
52 52 2
25 25
+ =
+ =
=
=
+ =
+ =
H
H
H
H
H
HH
H
H
H
H
L
L
L
L
2
50 2
50
25 2
5 2
72 2 122
49 2 1442 95
=
=
=
=
+ =
+ =
=
8.
==
+ =
+ =
=
=
=
+ =
95
122 2 132
144 2 1692 25
25
5
52 62
9.
10.
L
L
L
L
L
882
25 36 6461 64
2
+ == : false
not a right triangle
11. 22 42 62
4 16 3620 36
+ =+ =
= : false
not a right triangle
112. 62 82 102
36 64 100100 100
+ =+ =
= : true
is a right trriangle
Lesson Practice 18a - Lesson Practice 18B
soLUtionsGeoMetrY 167
Lesson Practice18A1.
2.
3.
4.
right
legs
hypotenuse
Pyythagorean
the hypotenuse squared;
if the leg
5.
squared plus the leg
squared equals the hypoteenuse
squared, the triangle is a
right triangle..
6.
7.
82 92 2
64 81 2
145 2
145
52 52 2
25 25
+ =
+ =
=
=
+ =
+ =
H
H
H
H
H
HH
H
H
H
H
L
L
L
L
2
50 2
50
25 2
5 2
72 2 122
49 2 1442 95
=
=
=
=
+ =
+ =
=
8.
==
+ =
+ =
=
=
=
+ =
95
122 2 132
144 2 1692 25
25
5
52 62
9.
10.
L
L
L
L
L
882
25 36 6461 64
2
+ == : false
not a right triangle
11. 22 42 62
4 16 3620 36
+ =+ =
= : false
not a right triangle
112. 62 82 102
36 64 100100 100
+ =+ =
= : true
is a right trriangle
Lesson Practice 18BLesson Practice18B1.
2.
3.
4.
90º
legs
hypotenuse
Pythhagorean theorem
the hypotenuse squared, then5.
the triangle is a right triangle. if
a trianglle is a right triangle, then
one leg squared pllus the other
leg squared equals the
hypotenuse squared.
726.
7.
+ =
+ =
=
=
+
102 2
49 100 2
149 2
149
32 3
H
H
H
H
22 2
9 9 2
18 2
18
9 2
3 2
242 2 252
576 2
=
+ =
=
=
=
=
+ =
+ =
H
H
H
H
H
H
L
L
8.
66252 49
49
7
5 22 2 10 3
2
5 5 2 2 2 10
L
L
L
L
L
=
=
=
( ) + = ( )( ) ( ) + =
9.
(( ) ( )
+ =
( ) + = ( )
+ =
=
10 3 3
25 4 2 100 9
25 2 2 100 3
50 2 3002
L
L
L
L 2250
250
25 10
5 10
42 52 62
16 25 3641 36
L
L
L
=
=
=
+ =+ =
=
10.
: faalse
not a right triangle
10211. + =+ =
242 262
100 576 6676676 676
162
=
+ =
: true
is a right triangle
12212. 2202
144 256 400400 400
+ == : true
is a right triangle
Lesson Practice 18B - sYsteMatic reVieW 18D
soLUtions GeoMetrY168
Lesson Practice18B1.
2.
3.
4.
90º
legs
hypotenuse
Pythhagorean theorem
the hypotenuse squared, then5.
the triangle is a right triangle. if
a trianglle is a right triangle, then
one leg squared pllus the other
leg squared equals the
hypotenuse squared.
726.
7.
+ =
+ =
=
=
+
102 2
49 100 2
149 2
149
32 3
H
H
H
H
22 2
9 9 2
18 2
18
9 2
3 2
242 2 252
576 2
=
+ =
=
=
=
=
+ =
+ =
H
H
H
H
H
H
L
L
8.
66252 49
49
7
5 22 2 10 3
2
5 5 2 2 2 10
L
L
L
L
L
=
=
=
( ) + = ( )( ) ( ) + =
9.
(( ) ( )
+ =
( ) + = ( )
+ =
=
10 3 3
25 4 2 100 9
25 2 2 100 3
50 2 3002
L
L
L
L 2250
250
25 10
5 10
42 52 62
16 25 3641 36
L
L
L
=
=
=
+ =+ =
=
10.
: faalse
not a right triangle
10211. + =+ =
242 262
100 576 6676676 676
162
=
+ =
: true
is a right triangle
12212. 2202
144 256 400400 400
+ == : true
is a right triangle
Systematic Review 18CSystematic Review 18C1. 72 102 49 100 149
149
+ = + =
is bbetween 12 and 13
any answer that is close
is accceptable.
2.
3.
72 102 2
72 102 2
49 100 2
149
+ =
+ =
+ =
=
H
H
H
H22
149
149 149
42 32 1
=
=
− =
H
4.
5.
:
cannot be simplified
66 9 7
7
32 2 42
32 2 42
9
− =
+ =
+ =
is between 2 and 3
6.
7.
L
L
++ =
=
=
+
L
L
L
no
2 162 7
7
7 2 65
52 72 92
8.
9.
10.
11.
≈
≠
.
hypotenuuse
12.
13.
a bh in
a r
= = ( ) ( ) =
= ( )
12
12
7 9 31 5 2
2 3 14 2
.
.π ≈ 22 12 56 2
31 5 12 56 18 94 2
11 5 7
( ) =
= − =
+
.
. . .
in
a in14.
15. 55 11 7 5 18 5
9 2 5 2 9 5 2 2
45 4 45 2 90
= +( ) =
( )( ) = ( ) ( ) =
= ( ) =
16.
117.
18.
9 200
12 2
3 1004
3 104
304
7 12
7 5
12
= =( )
=
=
=
.or
a bhh
in
a in
= ( )( ) =
( )( ) =
= ( ) ( ) =
12
4 2 3
2 2 3 4 3 2
5 4 3 20 3 219.
220. P in= ( ) ( ) =5 4 20
Systematic Review 18C1. 72 102 49 100 149
149
+ = + =
is bbetween 12 and 13
any answer that is close
is accceptable.
2.
3.
72 102 2
72 102 2
49 100 2
149
+ =
+ =
+ =
=
H
H
H
H22
149
149 149
42 32 1
=
=
− =
H
4.
5.
:
cannot be simplified
66 9 7
7
32 2 42
32 2 42
9
− =
+ =
+ =
is between 2 and 3
6.
7.
L
L
++ =
=
=
+
L
L
L
no
2 162 7
7
7 2 65
52 72 92
8.
9.
10.
11.
≈
≠
.
hypotenuuse
12.
13.
a bh in
a r
= = ( ) ( ) =
= ( )
12
12
7 9 31 5 2
2 3 14 2
.
.π ≈ 22 12 56 2
31 5 12 56 18 94 2
11 5 7
( ) =
= − =
+
.
. . .
in
a in14.
15. 55 11 7 5 18 5
9 2 5 2 9 5 2 2
45 4 45 2 90
= +( ) =
( )( ) = ( ) ( ) =
= ( ) =
16.
117.
18.
9 200
12 2
3 1004
3 104
304
7 12
7 5
12
= =( )
=
=
=
.or
a bhh
in
a in
= ( )( ) =
( )( ) =
= ( ) ( ) =
12
4 2 3
2 2 3 4 3 2
5 4 3 20 3 219.
220. P in= ( ) ( ) =5 4 20
Systematic Review 18DSystematic Review 18D1. 72 72 49 49 98
98
+ = + =
is betweeen 9 and 10
2.
3.
72 72 2
72 72 2
49 49 2
98 2
9
+ =
+ =
+ =
=
B
B
B
B
88
98 49 2 7 2
182 9 32
324 81 9
324 81 3 32
=
= =
− ( ) = − =
− ( ) =
B
4.
5.
44 243 81
81 9
182 9 32 2
182 9 32 2
324 81
− =
=
− ( ) =
− ( ) =
−
6.
7.
M
M
99 2
324 81 3 2
324 243 2
81 2
9
9 3 15 59
=
− ( ) =
− =
==
M
M
M
MM
8.
9.
≈ .
yyes
10.
11.
242 102 262
576 100 676676 676
+ =+ =
= : true
equuilateral
yes: they are both radii of the ci12. rrcle
13.
14.
50
180
º
ºm oaB m oBa m aoB
m oaB m oBa
∠ + ∠ + ∠ =
∠ + ∠ ++ =
∠ + ∠ =
∠ = =
+
50 180
130
130 2 65
3
º º
º
º º
m oaB m oBa
m oaB ÷
15. 33 2 3
6 24 144 12
2 15
5
2 31
2 3
12
=
( ) −( ) = − = −
= =
=
16.
17.
18. a bhh in= ( )( ) =
( ) =
12
12 5 6 30 6 2
30 6 6 180 6
6 triangles
19. iin
P in
2
6 12 7220. = ( ) =
sYsteMatic reVieW 18D - Lesson Practice 19a
soLUtionsGeoMetrY 169
Systematic Review 18D1. 72 72 49 49 98
98
+ = + =
is betweeen 9 and 10
2.
3.
72 72 2
72 72 2
49 49 2
98 2
9
+ =
+ =
+ =
=
B
B
B
B
88
98 49 2 7 2
182 9 32
324 81 9
324 81 3 32
=
= =
− ( ) = − =
− ( ) =
B
4.
5.
44 243 81
81 9
182 9 32 2
182 9 32 2
324 81
− =
=
− ( ) =
− ( ) =
−
6.
7.
M
M
99 2
324 81 3 2
324 243 2
81 2
9
9 3 15 59
=
− ( ) =
− =
==
M
M
M
MM
8.
9.
≈ .
yyes
10.
11.
242 102 262
576 100 676676 676
+ =+ =
= : true
equuilateral
yes: they are both radii of the ci12. rrcle
13.
14.
50
180
º
ºm oaB m oBa m aoB
m oaB m oBa
∠ + ∠ + ∠ =
∠ + ∠ ++ =
∠ + ∠ =
∠ = =
+
50 180
130
130 2 65
3
º º
º
º º
m oaB m oBa
m oaB ÷
15. 33 2 3
6 24 144 12
2 15
5
2 31
2 3
12
=
( ) −( ) = − = −
= =
=
16.
17.
18. a bhh in= ( )( ) =
( ) =
12
12 5 6 30 6 2
30 6 6 180 6
6 triangles
19. iin
P in
2
6 12 7220. = ( ) =
Systematic Review 18ESystematic Review 18E
1. 62 2 32
36 4 9
36 4 3
+ ( ) = + ( ) =
+ ( ) == + =
= + ( )
36 12 48
48
2 62 2 32
is between 6 and 7
2.
3.
r
r22 62 2 32
2 36 4 9
2 36 4 3
2 36 122 48
4
= + ( )= + ( )= + ( )
= +
=
=
r
r
r
r
r 88
48 16 3 4 3
112 112 121 121 242
242
4.
5.
= =
+ = + =
is betweenn 15 and 16
6.
7.
t
t
t
t
2 112 112
2 112 112
2 121 121
= +
= +
= +22 242
242 121 2 11 2
11 2 15 56
2 72
2 3
=
= = =
⊕
( ) + ( )
t
yes
8.
9.
.
22
2 102
4 49 4 9 4 100
4 7 4 3 4 10
28 12 4040 4
= ( )+ =
( ) + ( ) = ( )+ =
= 00: true
the sum of the legs squared
equals t
10.
hhe hypotenuse squared.
es: it would be a ri11. y gght triangle
with equal legs. 45 45 90º º º− −( )
12. VV Bh r h
m
V Bh r
= = ( ) +( ) ( )
=
= =
π ≈
π
2 3 14 2 32
12
942 3
2
.
13. hh
m
V
≈ 3 14 22 12
150 72 3
942 150 72
.
.
.
( ) ( )( )
=
= − =14. 7791 28 3
5 2 4 3 5 2 4 3
. m
15.
1
− = − :
cannot be simplified
66. 5 6 2 10
5 2 6 10 10 60
10 4 15 10 2 15 20 1
( )( ) =
( ) ( ) = =
= ( )( ) = 55
100
25
41
21
2
100
25
105
2
2 3 10
17.
18.
= = =
= =
+ − = +
or
X X X
:
55 2
2 2 3 3 1
2 6 3
( ) −( )
− − = −( ) +( )
+ − = +( ) −
X
X X X X
X X X X
19.
20. 22( )
Systematic Review 18E
1. 62 2 32
36 4 9
36 4 3
+ ( ) = + ( ) =
+ ( ) == + =
= + ( )
36 12 48
48
2 62 2 32
is between 6 and 7
2.
3.
r
r22 62 2 32
2 36 4 9
2 36 4 3
2 36 122 48
4
= + ( )= + ( )= + ( )
= +
=
=
r
r
r
r
r 88
48 16 3 4 3
112 112 121 121 242
242
4.
5.
= =
+ = + =
is betweenn 15 and 16
6.
7.
t
t
t
t
2 112 112
2 112 112
2 121 121
= +
= +
= +22 242
242 121 2 11 2
11 2 15 56
2 72
2 3
=
= = =
⊕
( ) + ( )
t
yes
8.
9.
.
22
2 102
4 49 4 9 4 100
4 7 4 3 4 10
28 12 4040 4
= ( )+ =
( ) + ( ) = ( )+ =
= 00: true
the sum of the legs squared
equals t
10.
hhe hypotenuse squared.
es: it would be a ri11. y gght triangle
with equal legs. 45 45 90º º º− −( )
12. VV Bh r h
m
V Bh r
= = ( ) +( ) ( )
=
= =
π ≈
π
2 3 14 2 32
12
942 3
2
.
13. hh
m
V
≈ 3 14 22 12
150 72 3
942 150 72
.
.
.
( ) ( )( )
=
= − =14. 7791 28 3
5 2 4 3 5 2 4 3
. m
15.
1
− = − :
cannot be simplified
66. 5 6 2 10
5 2 6 10 10 60
10 4 15 10 2 15 20 1
( )( ) =
( ) ( ) = =
= ( )( ) = 55
100
25
41
21
2
100
25
105
2
2 3 10
17.
18.
= = =
= =
+ − = +
or
X X X
:
55 2
2 2 3 3 1
2 6 3
( ) −( )
− − = −( ) +( )
+ − = +( ) −
X
X X X X
X X X X
19.
20. 22( )
Lesson Practice 19ALesson Practice19A1.
2.
9
2
9 2
2 2
9 2
4
9 22
10
5
10 5
5 5
= = =
= == =
= =
= = =
= =
10 5
25
10 55
2 51
2 5
6
3
6 3
3 3
6 3
9
6 33
2 31
2 3
8
3
3.
4. == = =
= = =
= =
8 3
3 3
8 3
9
8 33
15
5
15 5
5 5
15 5
25
15 55
3 51
3 5
9
3
5.
6. == = =
= =
+ = + =
+
9 3
3 3
9 3
9
9 33
3 31
3 3
3
2
6
5
3 2
2 2
6 5
5 5
3 2
4
6 5
25
7.
== + =
( )( ) +
( )( ) = + =
+
3 22
6 55
3 2 52 5
6 5 25 2
15 210
12 510
15 2 122 510
: note that although 10
and 15 have a commoon factor, and 10
and 12 have a common factor, there is
no factor that is common to all threee
terms, so this fraction cannot be
reduced.
seee the next solution for an example
of one thatt can be reduced.
Lesson Practice 19a - Lesson Practice 19a
soLUtions GeoMetrY170
8. 4
5
2
6
4 5
5 5
2 6
6 6
4 5
25
2 6
36
4 55
2 66
4 5 65 6
2
+ = + =
+ = + =
( )( ) + 66 5
6 524 5
3010 6
30
24 5 10 630
2 12 5 5 62 15
( )( ) = + =
+ =+( )
( ) =
112 5 5 615
4
2
9
5
4 2
2 2
9 5
5 5
4 2
4
9 5
25
4 22
9 55
4 2 5
+
+ = + =
+ = + =
9.
(( )( ) +
( )( ) = + =
+ =+
2 59 5 2
5 220 2
1018 5
10
20 2 18 510
2 10 2 9 5(( )( ) =
+
2 5
10 2 9 55
10. 5
10
3
8
5 10
10 10
3 8
8 8
5 10
100
3 8
64
5 1010
3 88
10
− = − =
− = − =
223 8
810 42 4
3 88
4 108
3 88
4 10 3 88
4 10 3 4 28
− =( )
( ) − =
− = − =
− = 44 10 3 2 28
2 2 10 3 2
2 42 10 3 2
4
5
7
6
2
5 7
7 7
− ( )=
−( )( ) = −
+ = +11. 66 2
2 2
5 7
49
6 2
4
5 77
6 22
5 7 27 2
6 2 72 7
10 71
=
+ = + =
( )( ) +
( )( ) =
4442 2
1410 7 42 2
14
2 5 7 21 22 7
5 7 21 27
5 2
3
+ = + =
+( )( ) = +
−12. 44 3
5
5 2 3
3 3
4 3 5
5 5
5 6
9
4 15
25
5 63
4 155
5 6 53 5
4
= − =
− = − =
( )( ) − 115 3
5 3
25 615
12 1515
25 6 12 1515
( )( ) =
− = −
13. denominatorr
one
denominator; common
14.
15.
Lesson Practice19A1.
2.
9
2
9 2
2 2
9 2
4
9 22
10
5
10 5
5 5
= = =
= == =
= =
= = =
= =
10 5
25
10 55
2 51
2 5
6
3
6 3
3 3
6 3
9
6 33
2 31
2 3
8
3
3.
4. == = =
= = =
= =
8 3
3 3
8 3
9
8 33
15
5
15 5
5 5
15 5
25
15 55
3 51
3 5
9
3
5.
6. == = =
= =
+ = + =
+
9 3
3 3
9 3
9
9 33
3 31
3 3
3
2
6
5
3 2
2 2
6 5
5 5
3 2
4
6 5
25
7.
== + =
( )( ) +
( )( ) = + =
+
3 22
6 55
3 2 52 5
6 5 25 2
15 210
12 510
15 2 122 510
: note that although 10
and 15 have a commoon factor, and 10
and 12 have a common factor, there is
no factor that is common to all threee
terms, so this fraction cannot be
reduced.
seee the next solution for an example
of one thatt can be reduced.
8. 4
5
2
6
4 5
5 5
2 6
6 6
4 5
25
2 6
36
4 55
2 66
4 5 65 6
2
+ = + =
+ = + =
( )( ) + 66 5
6 524 5
3010 6
30
24 5 10 630
2 12 5 5 62 15
( )( ) = + =
+ =+( )
( ) =
112 5 5 615
4
2
9
5
4 2
2 2
9 5
5 5
4 2
4
9 5
25
4 22
9 55
4 2 5
+
+ = + =
+ = + =
9.
(( )( ) +
( )( ) = + =
+ =+
2 59 5 2
5 220 2
1018 5
10
20 2 18 510
2 10 2 9 5(( )( ) =
+
2 5
10 2 9 55
Lesson Practice 19B - sYsteMatic reVieW 19c
soLUtionsGeoMetrY 171
10. 8 3
2
5 6
5
8 3 2
2 2
5 6 5
5 5
8 6
4
5 30
25
8 62
5 305
8 6 5
− = − =
− = − =
( )22 5
5 30 25 2
40 610
10 3010
40 6 10 3010
10 4 6 3
( ) −( )
( ) = − =
− =− 00
10 1
4 6 30
3
5
7
2
3 5
5 5
7 2
2 2
3 5
25
7 2
4
3 55
( )( ) =
−
+ = + =
+ = +
11.
77 22
3 5 25 2
7 2 52 5
6 510
35 210
6 5 35 210
=
( )( ) +
( )( ) = + =
+
12. 4 11
3
2 5
7
4 11 3
3 3
2 5 7
7 7
4 33
9
2 35
49
4 333
2 357
4
+ = + =
+ = + =
333 73 7
2 35 37 3
28 3321
6 3521
28 33 6 3521
( )( ) +
( )( ) = + =
+
13..
14.
15.
one
radical or square root
common denominaator
Systematic Review 19CSystematic Review 19C1.
2.
6
7
6 7
7 7
6 7
49
6 77
8
2
8 2
2
= = =
=22
8 2
4
8 22
4 21
4 2
6 2
3
6 2 3
3 3
6 6
9
6 63
2 6
2 3 6 2
= = =
=
= = = =
+ =
3.
4. 22 3 6 2
4 3 7 15
4 7 3 1
+
( )( ) =
( )( )
cannot be simplified
5.
55 28 45
28 9 5 28 3 5 84 5
36
6
61
6
= =
= ( ) =
= =6.
Lesson Practice 19BLesson Practice19B1.
2.
11 5
5
111
11
18
2
18 2
2 2
18 2
= =
= =44
18 22
9 21
9 2
12
6
12 6
6 6
12 6
36
12 66
2 61
2 6
7
2
= =
=
= = = =
=
3.
4. == = =
= = = =
=
7 2
2 2
7 2
4
7 22
6
3
6 3
3 3
6 3
9
6 33
2 31
2 3
5.
6. 9 6
5
9 6 5
5 5
9 30
25
9 305
= = =
7. 4 7
2
3 7
2
4 7 3 7
2
7 7
2
7 7 2
2 2
7 14
4
7 142
+ = + = =
= =
8. 129
184
4 39
9 24
2 39
3 24
2 3 49 4
3 2 94 9
8
− = − =
− =( )
( ) −( )
( ) =
3336
27 236
8 3 27 236
5
2
7
8
5 2
2 2
7 8
8 8
5 2
4
7 8
64
5
− = −
+ = + =
+ =
9.
222
7 88
5 22
7 4 28
5 22
7 2 28
5 2 22 2
7 24
10 24
+ =
+ = +( )
=
( )( ) + = + 77 2
417 2
4=
sYsteMatic reVieW 19c - sYsteMatic reVieW 19D
soLUtions GeoMetrY172
Systematic Review 19C1.
2.
6
7
6 7
7 7
6 7
49
6 77
8
2
8 2
2
= = =
=22
8 2
4
8 22
4 21
4 2
6 2
3
6 2 3
3 3
6 6
9
6 63
2 6
2 3 6 2
= = =
=
= = = =
+ =
3.
4. 22 3 6 2
4 3 7 15
4 7 3 1
+
( )( ) =
( )( )
cannot be simplified
5.
55 28 45
28 9 5 28 3 5 84 5
36
6
61
6
= =
= ( ) =
= =6.
7. 10 10
7
2 6
11
10 10 7
7 7
2 6 11
11 11
10 70
49
2 66
121
10 7
− =
− =
− = 007
2 6611
10 70 117 11
2 66 711 7
110 7077
14 6
− =
( )( ) −
( )( ) =
− 6677
110 70 14 6677
24 13
3
3 2
3
24 13 3
3 3
3 2 3
3 3
24 39
=
−
+ = + =8.
99
3 6
9
24 393
3 63
24 39 3 63
3 8 39 6
3 1
8 39 6
+ = + =
+ =+( )
( ) =
+
9.
10.
52 102 25 100 125
125
5
+ = + =
is between 11 and 12
22 102 2
52 102 2
25 100 2
125 2
125 25 5 5
+ =
+ =
+ =
=
= = =
Q
Q
Q
Q
Q
11.
55
2 162 202
2 256 40
12.
13.
guess about 12
X
X
+ =
+ = 002 144
12XX
==
14. X
X
XX
2 162 202
2 256 4002 144
12
+ =
+ =
==
15. V Bh r h
i
= =
( ) ( ) ( ) =
13
13
2
13
3 14 112 14 3 1 811 05
π ≈
. . , . nn3
16.
17.
check with ruler and protractor
the meassures of the angles of a
quadrilateral add up tto 360º. in a
rhombus, opposite angles are
conggruent, because they are
formed by transversalss cutting
parallel lines. if two of the angles
mmeasure 60º, then the other two
would be:
360º − 22 60
360 120 240
×( ) =
− =
º
º º º
if they add up to 240º, aand are
equal, then each must have a
measure off:
8 tri
240 2 120
12
12
5 3 2 7 5 2 2
º º
.
÷ =
= = ( ) ( ) =18. a bh in
aangles
19.
20.
8 7 5 2 60 2 2
8 5 40
( ) ( ) =
= ( ) =
. in
P in
Systematic Review 19DSystematic Review 19D1.
2.
9
5
9 5
5 5
9 5
25
9 55
6
2
6 2
2
= = =
=22
6 2
4
6 22
3 2
5 10
6
5 10 6
6 6
5 60
36
5 606
5 4 156
5 2
= = =
= = = =
=(
3.
))=
+ = +
156
5 153
5 6 2 10 5 6 2 104. :
cannot be simplified
55.
6.
3 8 2 5 3 2 8 5 6 40
6 4 10 6 2 10 12 10
2 14
7
( )( ) = ( ) ( ) = =
= ( ) =
== =2 21
2 2
sYsteMatic reVieW 19D - sYsteMatic reVieW 19e
soLUtionsGeoMetrY 173
7. 2 5
6
2 2
3
2 5 6
6 6
2 2 3
3 3
2 30
36
2 6
9
2 306
2 63
303
2 6
+ = + =
+ = + =
+33
30 2 63
5 11
2
3 5
2
5 11 2
2 2
3 5 2
2 2
5 22
4
3 10
4
5 222
= +
− = − =
− =
8.
−− =
−
+ =
3 102
5 22 3 102
92 112 29. X
10.
11.
92 112 2
81 121 2
202 2
202 14 21
+ =
+ =
=
=
X
X
X
X units≈ .
aa bh units
L
L
= = ( ) ( ) =
+ =
+
12
12
9 11 49 5 2
2 52 132
2
.
12.
13. 552 132
2 25 1692 144
12
12
12
5
=
+ =
==
= = ( )
L
LL
a bh
units
14. 112 30 2( ) = units
15.
16.
check with protractor
45º: ccheck with protractor
17. a = ( ) × ( ) ×
( )
12
7 12
5
3 5 2
π ≈
. .. . .
. .
5 3 14 27 48 2
2 3 14 32
28 26
( ) ( )
= ( ) ( ) =
≈
π ≈
m
a r i18. nn
a
2
90360
14
14
28 26 25 28 26 7 0
19.
20.
ºº
. . . .
=
= ( ) = ( )( ) ≈ 77 2in
Systematic Review 19ESystematic Review 19E1.
2.
2 6
2 5
6
5
6 5
5 5
30
25
305
5
= = = =
110
1
2
2
2 2
2
4
22
12 13
2 13
122
6
8 2 2 14 8 2 2 14
= = = =
= =
+ = +
3.
4. :
cannot be simplified
5.
6.
2 7 5 8 2 5 7 8
10 56 10 4 14
10 2 14 20 14
14
( )( ) = ( ) ( ) =
= =
( ) =
77
14 7
7 7
14 7
49
14 77
2 71
2 7
15
5
20
2
15 5
5 5
20 2
2 2
= = = =
=
+ = +7. ==
+ = + =
+
15 5
25
20 2
4
15 55
20 22
3 5 10 2
8. 5 2
14
3 2
14
5 2 3 2
14
2 2
14
2 2 14
14 14
2 28
196
2 2814
− = − = =
= = =
2287
4 77
2 77
52 62 2
52 62 2
25 36 2
61 2
6
= =
+ =
+ =
+ =
=
9.
10.
X
X
X
X
11 2 7 81
12
5 6 15
2 2
= ⊕
= ( ) ( ) =
+
X units
a
B B
.
11.
12.
units2
(( ) =
+ ( ) =
+ ( )( ) =
+ =
2 2
2 22 2
2 2 2 2
2 4 2 2
5 2
X
B B X
B B B X
B B X
B
13.
==
=
= =
= = ( )( ) =
X
B X
X B B units
a bh B B B u
2
5 2
2 5 5
12
12
2 214. nnits
a r in
2
2 3 14 42 50 24 215.
16.
= ( ) ( ) =π ≈ . .
sector is 45º360º
of the circle=
= ( )
=
18
18
50 24
12
17. a .
. 55 50 24 6 28 2
9
12 9 3
1 000
23 1 0003
( ) ( )
= =
=
. .
, ,
≈ in
18.
19. = =
= ( ) = =
2102 100
4
32 4
323 820.
sYsteMatic reVieW 19e - sYsteMatic reVieW 20c
soLUtions GeoMetrY174
8. 5 2
14
3 2
14
5 2 3 2
14
2 2
14
2 2 14
14 14
2 28
196
2 2814
− = − = =
= = =
2287
4 77
2 77
52 62 2
52 62 2
25 36 2
61 2
6
= =
+ =
+ =
+ =
=
9.
10.
X
X
X
X
11 2 7 81
12
5 6 15
2 2
= ⊕
= ( ) ( ) =
+
X units
a
B B
.
11.
12.
units2
(( ) =
+ ( ) =
+ ( )( ) =
+ =
2 2
2 22 2
2 2 2 2
2 4 2 2
5 2
X
B B X
B B B X
B B X
B
13.
==
=
= =
= = ( )( ) =
X
B X
X B B units
a bh B B B u
2
5 2
2 5 5
12
12
2 214. nnits
a r in
2
2 3 14 42 50 24 215.
16.
= ( ) ( ) =π ≈ . .
sector is 45º360º
of the circle=
= ( )
=
18
18
50 24
12
17. a .
. 55 50 24 6 28 2
9
12 9 3
1 000
23 1 0003
( ) ( )
= =
=
. .
, ,
≈ in
18.
19. = =
= ( ) = =
2102 100
4
32 4
323 820.
Lesson Practice 20ALesson Practice 20A1.
2.
3.
90º; 45º
isosceles
hypoteenuse
equal4.
5. Pythagorean 6.
7.
8.
9.
10.
2
7 2
7
5 2
5 11.
12.
13.
14.
3 2
3 2 2 3 4 3 2 6
8 2
2
81
8
8
= = ( ) =
= =
15. since aBcD is a square, all four
sides are eqqual in length.
therefore, aBD has two
congrue
nnt sides, and is
a 45º − −45 90
10 2
º º .triangle
Lesson Practice 20BLesson Practice 20B1.
2.
3.
4.
5.
false
false
true
true
ffalse
true6.
7.
8.
9.
10.
11.
12.
8 2
8
6 3 2 6 6
6 3
5 2
2
51
5
5
=
= =
113.
14.
15.
10
2
10 2
2 2
10 2
4
10 22
5 21
5 2
5 2
14
2
14 2
2 2
= = =
= =
= == = =
=
14 2
4
14 22
7 21
7 2 cm
Systematic Review 20CSystematic Review 20C1.
2.
3.
4.
5.
5 2
5
20 2
2
201
20
20
= =
HH
H
H
H
2 32 72
2 9 492 58
58
= +
= +
=
=
sYsteMatic reVieW 20c - sYsteMatic reVieW 20D
soLUtionsGeoMetrY 175
Systematic Review 20C1.
2.
3.
4.
5.
5 2
5
20 2
2
201
20
20
= =
HH
H
H
H
2 32 72
2 9 492 58
58
= +
= +
=
=
6. 11 40 9 5 11 9 40 5
99 200 99 100 2
99 10 2 99
( )( ) = ( ) ( )= = =
( ) = 00 2
2 6 10 8 2 10 6 8
20 48 20 16 3
20 4 3 80 3
7. ( )( ) = ( ) ( ) =
= =
( ) =
8. 5 6
7
5
6
5 6 7
7 7
5 6
6 6
5 42
49
30
36
5 427
306
5 42 67
− = − =
− = − =
( )66
30 76 7
30 4242
7 3042
30 42 7 3042
( ) −( )
( ) =
− = −
9. check wiith ruler and protractor
10. 360 2 123
360
º º
º
− ×( ) =
−2246 114
114 2 57
º º
º º
=
=÷
11. Figure is a parallelogramm
a bh cm
n
= = ( )( ) =
−( ) => ( ) −( )6 3 5 21 2
2 180 5 2 180
.
º º12. ==
( )( ) =
=
3 180 540
540 5 108
º º
º º
total
sides÷ per sidee
13.
14.
a r
mm
c r
= ( )( )=
=
π ≈
π ≈
2 3 14 622
12 070 16 2
2 2
.
, .(( )( )( )
=
= ( )( )
3 14 62
389 36
43
3 43
3 14 13
.
.
.
mm
V r15. π ≈ ≈ 44 19 3
120360
13
2 3 14 32
.
ºº
.
m
a r
16.
17.
=
= ( )
circle:
π ≈ (( ) =
= ( ) =
28 26 2
13
28 26 9 42 2
.
. .
units
a unitssector:
118.
19.
a bh in= = ( )( ) =
(
12
12
8 4 7 16 7 2
5 16 7
5 triangles
)) =
= ( ) =
80 7 2
5 8 40
in
P in20.
8. 5 6
7
5
6
5 6 7
7 7
5 6
6 6
5 42
49
30
36
5 427
306
5 42 67
− = − =
− = − =
( )66
30 76 7
30 4242
7 3042
30 42 7 3042
( ) −( )
( ) =
− = −
9. check wiith ruler and protractor
10. 360 2 123
360
º º
º
− ×( ) =
−2246 114
114 2 57
º º
º º
=
=÷
11. Figure is a parallelogramm
a bh cm
n
= = ( )( ) =
−( ) => ( ) −( )6 3 5 21 2
2 180 5 2 180
.
º º12. ==
( )( ) =
=
3 180 540
540 5 108
º º
º º
total
sides÷ per sidee
13.
14.
a r
mm
c r
= ( )( )=
=
π ≈
π ≈
2 3 14 622
12 070 16 2
2 2
.
, .(( )( )( )
=
= ( )( )
3 14 62
389 36
43
3 43
3 14 13
.
.
.
mm
V r15. π ≈ ≈ 44 19 3
120360
13
2 3 14 32
.
ºº
.
m
a r
16.
17.
=
= ( )
circle:
π ≈ (( ) =
= ( ) =
28 26 2
13
28 26 9 42 2
.
. .
units
a unitssector:
118.
19.
a bh in= = ( )( ) =
(
12
12
8 4 7 16 7 2
5 16 7
5 triangles
)) =
= ( ) =
80 7 2
5 8 40
in
P in20.
Systematic Review 20DSystematic Review 20D1.
2.
3.
7 2 2 7 4 7 2 14
7 2
25
2
= = ( ) =
= 225 2
2 2
25 2
4
25 22
25 22
2 62 82
2 36 642 100
= =
= +
= +
=
4.
5. H
H
HH == 10
6. 12 3 4 18 12 4 3 18
48 54 48 9 6 48 3 6
144 6
( )( ) = ( )( ) =
= = ( ) =
7. 4 5 20 2 4 20 5 2
80 10
( )( ) = ( ) ( ) =
8. 2 25
5
5
25
2 5
5
1
510
5
1
5
11
5
11 5
5 5
11 5
25
11 55
+ =( )
+ =
+ = =
= =
99. check with ruler and protractor:
second pair of sides should
be 78
. angles should be 90in ºº.
10. P = + = + =
+
2 1 34
78
2 74
78
2 148
78
( ) ( )
( = =
= =
=
) ( )2 218
428
214
5 14
43
3
in
V r11.
1
π
22. n −( ) => ( ) −( ) =
( ) =
2 180 8 2 180
6 180 1 080
º º
º , º total;
1,080º per angle÷
π ≈
8 135
13
13
2
13
3 1
=
= =
º
.
13. V Bh r h
44 8 32
12 4 894 1 3( ) ( ) ( ) =
= =
. . . cm
V Bh r
14.
15.
diameter
π 22
3 14 82 12 2 411 52 3
2 2 2
h
cm
sa r rh
≈
π π ≈
. , .( ) ( )( ) =
= +16.
22 3 14 82 2 3 14 8 12
401 92 602 88 1
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+ =
. .
. . ,, .
ºº
.
004 8 2
60360
16
2 3 14
cm
a r
17.
18.
=
= ( )circle:
π ≈ 442 50 24 2
16
50 24 8 37 2
( ) =
= ( ).
. .
cm
a cmsector:
s
≈
19. eemicircle:
recta
a r in= ( ) ( ) =12
2 12
3 14 32 14 13 2π ≈ . .
nngle:
total:
a bh in
a
= = ( ) ( ) =
= + =
9 6 54 2
14 13 54 68 13. . iin2
20. perimeter of semicircle is half the
circummference of the circle:
P r= ( ) ( ) ( ) (12
2 12
2 3 14 3π ≈ . )) =
( )= +
9 42
9 6
.
:
in
P
rectangle exterior lines only
++ =
+ =
9 24
9 42 24 33 42. .
in
intotal:
⇒
sYsteMatic reVieW 20D - sYsteMatic reVieW 20e
soLUtions GeoMetrY176
8. 2 25
5
5
25
2 5
5
1
510
5
1
5
11
5
11 5
5 5
11 5
25
11 55
+ =( )
+ =
+ = =
= =
99. check with ruler and protractor:
second pair of sides should
be 78
. angles should be 90in ºº.
10. P = + = + =
+
2 1 34
78
2 74
78
2 148
78
( ) ( )
( = =
= =
=
) ( )2 218
428
214
5 14
43
3
in
V r11.
1
π
22. n −( ) => ( ) −( ) =
( ) =
2 180 8 2 180
6 180 1 080
º º
º , º total;
1,080º per angle÷
π ≈
8 135
13
13
2
13
3 1
=
= =
º
.
13. V Bh r h
44 8 32
12 4 894 1 3( ) ( ) ( ) =
= =
. . . cm
V Bh r
14.
15.
diameter
π 22
3 14 82 12 2 411 52 3
2 2 2
h
cm
sa r rh
≈
π π ≈
. , .( ) ( )( ) =
= +16.
22 3 14 82 2 3 14 8 12
401 92 602 88 1
( ) ( ) ( ) + ( ) ( ) ( ) ( ) =
+ =
. .
. . ,, .
ºº
.
004 8 2
60360
16
2 3 14
cm
a r
17.
18.
=
= ( )circle:
π ≈ 442 50 24 2
16
50 24 8 37 2
( ) =
= ( ).
. .
cm
a cmsector:
s
≈
19. eemicircle:
recta
a r in= ( ) ( ) =12
2 12
3 14 32 14 13 2π ≈ . .
nngle:
total:
a bh in
a
= = ( ) ( ) =
= + =
9 6 54 2
14 13 54 68 13. . iin2
20. perimeter of semicircle is half the
circummference of the circle:
P r= ( ) ( ) ( ) (12
2 12
2 3 14 3π ≈ . )) =
( )= +
9 42
9 6
.
:
in
P
rectangle exterior lines only
++ =
+ =
9 24
9 42 24 33 42. .
in
intotal:
Systematic Review 20ESystematic Review 20E1.
2.
3.
4
14 3 2 14 6
14 3
5 2
2
51
5
=
= =
..
5.
6.
5
2 42 102
2 16 1002 116
116 4 29 2 29
10 2
H
H
H
H
= +
= +
=
= = =
(( )( ) = ( ) ( ) =
= ( ) =
( )( ) = ( )
3 8 10 3 2 8
30 16 30 4 120
4 7 6 3 4 67. (( ) =
+ =( )
+ =
+ = + =
7 3 24 21
2 16
2
2
16
2 4
2
24
8
2
24
8 2
2 2
24
8 2
4
8.
++ = + =
( )( ) + = + =
24
8 22
24
8 2 22 2
24
16 24
24
17 24
9. Use your compass to draw a
circle. Draw the radius, andd use
your protractor to measure out
an angle oof 210º. see the end
of lesson 4 in the instrucction
anual for hints on drawing
or measuring
m
oobtuse angles.
prism
10+122
10.
11. a = ( )
= ( )
13
11 133 143
2 180 10 2 180
8 18
( ) =
−( ) => ( ) −( ) =
( )º º
units2
12. n
00 1 440
10 144
º , º
º
=
=
total;
1,440º per side
lin
÷
13. ee or segment or ray
angles are congruent14.
15. c == ( ) ( ) ( ) =2 2 3 14 6 37 68π ≈r in. .
16. Measure of minor aarc Bc
so arc is 14
of cir
=
∠ =
=
m Boc 90
90360
14
º
ºº
ccle
length of arc = ( ) =
−
14
37 68 9 42
37 68 9 4
. .
. .
in
17. 22 28 26
7 2 28 28
7 2 4 4
7 2
=
+ + =
( ) + +( ) =
( ) +( )
. in
X X
X X
X X
18.
++( )
+ − =
( ) + −( ) =
( ) +( ) −( )
2
3 2 15 18
3 2 5 6
3 6 1
19.
20
X X
X X
X X
.. 2 2 11 5
2 1 5
X X
X X
+ + =
+( ) +( )
Systematic Review 20E1.
2.
3.
4
14 3 2 14 6
14 3
5 2
2
51
5
=
= =
..
5.
6.
5
2 42 102
2 16 1002 116
116 4 29 2 29
10 2
H
H
H
H
= +
= +
=
= = =
(( )( ) = ( ) ( ) =
= ( ) =
( )( ) = ( )
3 8 10 3 2 8
30 16 30 4 120
4 7 6 3 4 67. (( ) =
+ =( )
+ =
+ = + =
7 3 24 21
2 16
2
2
16
2 4
2
24
8
2
24
8 2
2 2
24
8 2
4
8.
++ = + =
( )( ) + = + =
24
8 22
24
8 2 22 2
24
16 24
24
17 24
9. Use your compass to draw a
circle. Draw the radius, andd use
your protractor to measure out
an angle oof 210º. see the end
of lesson 4 in the instrucction
anual for hints on drawing
or measuring
m
oobtuse angles.
prism
10+122
10.
11. a = ( )
= ( )
13
11 133 143
2 180 10 2 180
8 18
( ) =
−( ) => ( ) −( ) =
( )º º
units2
12. n
00 1 440
10 144
º , º
º
=
=
total;
1,440º per side
lin
÷
13. ee or segment or ray
angles are congruent14.
15. c == ( ) ( ) ( ) =2 2 3 14 6 37 68π ≈r in. .
16. Measure of minor aarc Bc
so arc is 14
of cir
=
∠ =
=
m Boc 90
90360
14
º
ºº
ccle
length of arc = ( ) =
−
14
37 68 9 42
37 68 9 4
. .
. .
in
17. 22 28 26
7 2 28 28
7 2 4 4
7 2
=
+ + =
( ) + +( ) =
( ) +( )
. in
X X
X X
X X
18.
++( )
+ − =
( ) + −( ) =
( ) +( ) −( )
2
3 2 15 18
3 2 5 6
3 6 1
19.
20
X X
X X
X X
.. 2 2 11 5
2 1 5
X X
X X
+ + =
+( ) +( )
sYsteMatic reVieW 20e - sYsteMatic reVieW 21c
soLUtionsGeoMetrY 177
Systematic Review 20E1.
2.
3.
4
14 3 2 14 6
14 3
5 2
2
51
5
=
= =
..
5.
6.
5
2 42 102
2 16 1002 116
116 4 29 2 29
10 2
H
H
H
H
= +
= +
=
= = =
(( )( ) = ( ) ( ) =
= ( ) =
( )( ) = ( )
3 8 10 3 2 8
30 16 30 4 120
4 7 6 3 4 67. (( ) =
+ =( )
+ =
+ = + =
7 3 24 21
2 16
2
2
16
2 4
2
24
8
2
24
8 2
2 2
24
8 2
4
8.
++ = + =
( )( ) + = + =
24
8 22
24
8 2 22 2
24
16 24
24
17 24
9. Use your compass to draw a
circle. Draw the radius, andd use
your protractor to measure out
an angle oof 210º. see the end
of lesson 4 in the instrucction
anual for hints on drawing
or measuring
m
oobtuse angles.
prism
10+122
10.
11. a = ( )
= ( )
13
11 133 143
2 180 10 2 180
8 18
( ) =
−( ) => ( ) −( ) =
( )º º
units2
12. n
00 1 440
10 144
º , º
º
=
=
total;
1,440º per side
lin
÷
13. ee or segment or ray
angles are congruent14.
15. c == ( ) ( ) ( ) =2 2 3 14 6 37 68π ≈r in. .
16. Measure of minor aarc Bc
so arc is 14
of cir
=
∠ =
=
m Boc 90
90360
14
º
ºº
ccle
length of arc = ( ) =
−
14
37 68 9 42
37 68 9 4
. .
. .
in
17. 22 28 26
7 2 28 28
7 2 4 4
7 2
=
+ + =
( ) + +( ) =
( ) +( )
. in
X X
X X
X X
18.
++( )
+ − =
( ) + −( ) =
( ) +( ) −( )
2
3 2 15 18
3 2 5 6
3 6 1
19.
20
X X
X X
X X
.. 2 2 11 5
2 1 5
X X
X X
+ + =
+( ) +( )
Lesson Practice 21ALesson Practice 21A1.
2.
3.
4.
5.
60 90
3
;
scalene
2
Pythhagorean
6.
7.
8.
9.
2 3
5 3
5 2 10
6 3
3
61
; ;multiply
( ) =
= = 66
6 2 12
10 32
5 31
5 3
5 3 3 5 9 5 3 15
8
10.
11.
12.
13.
( ) =
= =
= = ( ) =
33
8 2 1614.
15.
( ) =
Because of the relationship betweeen the
lengths of the sides, we know that DB cc is
triangle, so the measures a 30 60 90º º º− −
aare 30º and 60º respectively.
Lesson Practice 21BLesson Practice 21B1.
2.
3.
4.
5
false
false
true
false
..
6.
7.
8.
9.
10.
true
true
7 3
3
71
7
7 2 14
8 32
4 31
4 3
4
= =
( ) =
= =
33 3 4 9 4 3 12
11 2 22
11 3
12 3
3
121
12
= = ( ) =
( ) =
= =
11.
12.
13.
14..
15.
12 2 24( ) =
=radius of circle
ypotenuse of trih aangle =
( ) =5 2 10 cm
Lesson Practice 21B1.
2.
3.
4.
5
false
false
true
false
..
6.
7.
8.
9.
10.
true
true
7 3
3
71
7
7 2 14
8 32
4 31
4 3
4
= =
( ) =
= =
33 3 4 9 4 3 12
11 2 22
11 3
12 3
3
121
12
= = ( ) =
( ) =
= =
11.
12.
13.
14..
15.
12 2 24( ) =
=radius of circle
ypotenuse of trih aangle =
( ) =5 2 10 cm
Systematic Review 21CSystematic Review 21C1.
2.
3.
7 2 14
7 3
9 3
32 9
12
( ) =
( ) = ( )) = ( ) =
= =
= = ( ) =
9 2 18
9 3
3
91
9
2 2 2 2 4 2 2 4
2 2
6 2
6
4.
5.
6.
7.
8.
9.. 10 20 2 7 10 2 20 7
20 140 20 4 35
20 2 35 40 3
( )( ) = ( ) ( ) =
= =
( ) = 55
3 5 4 5 3 4 5 7 5
4 8
4
5
16
4 82
54
4 8 22 2
10.
11.
+ = +( ) =
+ = + =
( )( ) ++ = + =
+ = + =
( ) + = +
=
54
8 84
54
8 8 54
8 4 2 54
8 2 2 54
16 2 54
2 212. H 222 162
2 484 2562 740
740 4 185 2 185
4
+
= +
=
= = =
=
H
H
H
sa r13. π 22 4 3 14 1 52
28 26 2
≈ ( ) ( ) ( ) =. .
. in
14.
15.
see drawing
extterior angle
sides
= − =
=
180 140 40
360 40 9
º º º
º º÷
16. c == ( ) ( ) ( ) =
=
2 2 3 14 7 43 96
180360
12
12
4
π ≈r in. .
ºº
17.
18. 33 96 21 98. .( ) = in
19. area of semicircle is half arrea
of circle:
t
a r
cm
= ( ) ( )12
2 12
3 14 2 52
9 81 2
π ≈
≈
. .
.
oop base of trapezoid is twice
the radius:
a = +5 822
3 19 5 2
9 81 19 5 29 31 2
total:
semi
( ) =
+ =
.
. . .
cm
cm
20. ccircle:
trape
c r
cm
= ( ) ( ) ( ) ( )
=
12
2 12
2 3 14 2 5
7 85
π ≈ . .
.
zzoid:
total:
P cm
cm
= + + =
+ =
4 5 8 4 16 5
7 85 16 5 24 35
. .
. . .
sYsteMatic reVieW 21c - sYsteMatic reVieW 21D
soLUtions GeoMetrY178
Systematic Review 21C1.
2.
3.
7 2 14
7 3
9 3
32 9
12
( ) =
( ) = ( )) = ( ) =
= =
= = ( ) =
9 2 18
9 3
3
91
9
2 2 2 2 4 2 2 4
2 2
6 2
6
4.
5.
6.
7.
8.
9.. 10 20 2 7 10 2 20 7
20 140 20 4 35
20 2 35 40 3
( )( ) = ( ) ( ) =
= =
( ) = 55
3 5 4 5 3 4 5 7 5
4 8
4
5
16
4 82
54
4 8 22 2
10.
11.
+ = +( ) =
+ = + =
( )( ) ++ = + =
+ = + =
( ) + = +
=
54
8 84
54
8 8 54
8 4 2 54
8 2 2 54
16 2 54
2 212. H 222 162
2 484 2562 740
740 4 185 2 185
4
+
= +
=
= = =
=
H
H
H
sa r13. π 22 4 3 14 1 52
28 26 2
≈ ( ) ( ) ( ) =. .
. in
14.
15.
see drawing
extterior angle
sides
= − =
=
180 140 40
360 40 9
º º º
º º÷
16. c == ( ) ( ) ( ) =
=
2 2 3 14 7 43 96
180360
12
12
4
π ≈r in. .
ºº
17.
18. 33 96 21 98. .( ) = in
19. area of semicircle is half arrea
of circle:
t
a r
cm
= ( ) ( )12
2 12
3 14 2 52
9 81 2
π ≈
≈
. .
.
oop base of trapezoid is twice
the radius:
a = +5 822
3 19 5 2
9 81 19 5 29 31 2
total:
semi
( ) =
+ =
.
. . .
cm
cm
20. ccircle:
trape
c r
cm
= ( ) ( ) ( ) ( )
=
12
2 12
2 3 14 2 5
7 85
π ≈ . .
.
zzoid:
total:
P cm
cm
= + + =
+ =
4 5 8 4 16 5
7 85 16 5 24 35
. .
. . .
Systematic Review 21C1.
2.
3.
7 2 14
7 3
9 3
32 9
12
( ) =
( ) = ( )) = ( ) =
= =
= = ( ) =
9 2 18
9 3
3
91
9
2 2 2 2 4 2 2 4
2 2
6 2
6
4.
5.
6.
7.
8.
9.. 10 20 2 7 10 2 20 7
20 140 20 4 35
20 2 35 40 3
( )( ) = ( ) ( ) =
= =
( ) = 55
3 5 4 5 3 4 5 7 5
4 8
4
5
16
4 82
54
4 8 22 2
10.
11.
+ = +( ) =
+ = + =
( )( ) ++ = + =
+ = + =
( ) + = +
=
54
8 84
54
8 8 54
8 4 2 54
8 2 2 54
16 2 54
2 212. H 222 162
2 484 2562 740
740 4 185 2 185
4
+
= +
=
= = =
=
H
H
H
sa r13. π 22 4 3 14 1 52
28 26 2
≈ ( ) ( ) ( ) =. .
. in
14.
15.
see drawing
extterior angle
sides
= − =
=
180 140 40
360 40 9
º º º
º º÷
16. c == ( ) ( ) ( ) =
=
2 2 3 14 7 43 96
180360
12
12
4
π ≈r in. .
ºº
17.
18. 33 96 21 98. .( ) = in
19. area of semicircle is half arrea
of circle:
t
a r
cm
= ( ) ( )12
2 12
3 14 2 52
9 81 2
π ≈
≈
. .
.
oop base of trapezoid is twice
the radius:
a = +5 822
3 19 5 2
9 81 19 5 29 31 2
total:
semi
( ) =
+ =
.
. . .
cm
cm
20. ccircle:
trape
c r
cm
= ( ) ( ) ( ) ( )
=
12
2 12
2 3 14 2 5
7 85
π ≈ . .
.
zzoid:
total:
P cm
cm
= + + =
+ =
4 5 8 4 16 5
7 85 16 5 24 35
. .
. . .
Systematic Review 21DSystematic Review 21D1.
2.
4 2
3
4 2 3
3 3
4 6
9
4 63
4 63
2
= = =
( )) =
= =
= =
= =
8 63
12 32
6 31
6 3
6 3
3
61
6
15 2
15
5 2 2 5 4
3.
4.
5.
6.
7. 55 2 10
5 2
4 4 2 4 4 2 4 4
8 16 8 4 32
2
( ) =
( )( ) = ( )( ) =
= ( ) =
8.
9.
10. 118 5 9 2 9 2 5 3
2 3 2 15 6 2 15
3 6
7
4 6
5
3 6 7
7 7
+ = + ( ) =
( ) + = +
+ = +11. 44 6 5
5 5
3 42
49
4 30
25
3 427
4 305
3 42 57 5
4 30 75
=
+ = + =
( )( ) +
( )77
15 4235
28 3035
15 42 28 3035
132 2 172
169
( ) =
+ =
+
+ =
+
12. L
LL
L
L
sa r
2 2892 120
120 4 30 2 30
4 2 4 3 14 2
=
=
= = =
= ( ) ( )13. π ≈ . (( )
=
2
50 24 2. in
14.
15.
see drawing
exterior anglee = − =
=
= ( )
180 150 30
360 30 12
2 2 3
º º º
º º
.
÷
π ≈
sides
c r16. 114 4 25 12
270360
34
34
25 12 75
( ) ( ) =
=
( ) =
.
ºº
. .
in
17.
18. (( ) ( ) =
=
25 12 18 84. . in
V area19. of base times heighht
triangles:
=
( ) ( ) ( ) =
=
12
6 2 9 1 13 366 73 3
2
. . . in
a
20.
(( ) = ( ) ( ) ( )
=
12
2 12
6 2 9 1
56 42 2
bh
in
. .
.
large rectanglees:
small rectangle:
a in
a
= ( )( ) ( ) =
= (
2 11 13 286 2
6 2. )) ( ) =
= + +
=
13 80 6 2
56 42 286 80 6
423 02
.
. .
.
in
a
total:
iin2
Systematic Review 21D1.
2.
4 2
3
4 2 3
3 3
4 6
9
4 63
4 63
2
= = =
( )) =
= =
= =
= =
8 63
12 32
6 31
6 3
6 3
3
61
6
15 2
15
5 2 2 5 4
3.
4.
5.
6.
7. 55 2 10
5 2
4 4 2 4 4 2 4 4
8 16 8 4 32
2
( ) =
( )( ) = ( )( ) =
= ( ) =
8.
9.
10. 118 5 9 2 9 2 5 3
2 3 2 15 6 2 15
3 6
7
4 6
5
3 6 7
7 7
+ = + ( ) =
( ) + = +
+ = +11. 44 6 5
5 5
3 42
49
4 30
25
3 427
4 305
3 42 57 5
4 30 75
=
+ = + =
( )( ) +
( )77
15 4235
28 3035
15 42 28 3035
132 2 172
169
( ) =
+ =
+
+ =
+
12. L
LL
L
L
sa r
2 2892 120
120 4 30 2 30
4 2 4 3 14 2
=
=
= = =
= ( ) ( )13. π ≈ . (( )
=
2
50 24 2. in
14.
15.
see drawing
exterior anglee = − =
=
= ( )
180 150 30
360 30 12
2 2 3
º º º
º º
.
÷
π ≈
sides
c r16. 114 4 25 12
270360
34
34
25 12 75
( ) ( ) =
=
( ) =
.
ºº
. .
in
17.
18. (( ) ( ) =
=
25 12 18 84. . in
V area19. of base times heighht
triangles:
=
( ) ( ) ( ) =
=
12
6 2 9 1 13 366 73 3
2
. . . in
a
20.
(( ) = ( ) ( ) ( )
=
12
2 12
6 2 9 1
56 42 2
bh
in
. .
.
large rectanglees:
small rectangle:
a in
a
= ( )( ) ( ) =
= (
2 11 13 286 2
6 2. )) ( ) =
= + +
=
13 80 6 2
56 42 286 80 6
423 02
.
. .
.
in
a
total:
iin2
sYsteMatic reVieW 21D - Lesson Practice 22a
soLUtionsGeoMetrY 179
Systematic Review 21D1.
2.
4 2
3
4 2 3
3 3
4 6
9
4 63
4 63
2
= = =
( )) =
= =
= =
= =
8 63
12 32
6 31
6 3
6 3
3
61
6
15 2
15
5 2 2 5 4
3.
4.
5.
6.
7. 55 2 10
5 2
4 4 2 4 4 2 4 4
8 16 8 4 32
2
( ) =
( )( ) = ( )( ) =
= ( ) =
8.
9.
10. 118 5 9 2 9 2 5 3
2 3 2 15 6 2 15
3 6
7
4 6
5
3 6 7
7 7
+ = + ( ) =
( ) + = +
+ = +11. 44 6 5
5 5
3 42
49
4 30
25
3 427
4 305
3 42 57 5
4 30 75
=
+ = + =
( )( ) +
( )77
15 4235
28 3035
15 42 28 3035
132 2 172
169
( ) =
+ =
+
+ =
+
12. L
LL
L
L
sa r
2 2892 120
120 4 30 2 30
4 2 4 3 14 2
=
=
= = =
= ( ) ( )13. π ≈ . (( )
=
2
50 24 2. in
14.
15.
see drawing
exterior anglee = − =
=
= ( )
180 150 30
360 30 12
2 2 3
º º º
º º
.
÷
π ≈
sides
c r16. 114 4 25 12
270360
34
34
25 12 75
( ) ( ) =
=
( ) =
.
ºº
. .
in
17.
18. (( ) ( ) =
=
25 12 18 84. . in
V area19. of base times heighht
triangles:
=
( ) ( ) ( ) =
=
12
6 2 9 1 13 366 73 3
2
. . . in
a
20.
(( ) = ( ) ( ) ( )
=
12
2 12
6 2 9 1
56 42 2
bh
in
. .
.
large rectanglees:
small rectangle:
a in
a
= ( )( ) ( ) =
= (
2 11 13 286 2
6 2. )) ( ) =
= + +
=
13 80 6 2
56 42 286 80 6
423 02
.
. .
.
in
a
total:
iin2
Systematic Review 21ESystematic Review 21E1.
2.
3.
3
31
1 2 2
16 7 32
16 21
=
( ) =
=22
8 211
8 21
16 72
8 71
8 7
8 2
8
7 2 2 7 4 7 2 14
= =
= =
= = ( ) =
4.
5.
6.
7.
88.
9.
10
7 2
3 10 7 10 3 7 10 10
21 100 21 10 210
( )( ) = ( ) ( ) =
= ( ) =
..
11.
5 7 4 3 5 7 4 3
10 18
2 6
10
+ = +
=
:
cannot be simplified
332
5 31
5 3
112 2 222
121 2 4842 363
363 12
= =
+ =
+ =
=
= =
12. L
L
L
L 11 3 11 3
12
12
11 11 3
60 5 3 2
=
= = ( )( ) =
.
units
a bh
units
13.
114. one triangle:
six
a bh
in
= = ( )( ) =12
12
5 11 14
35 11 2
triangles:
a in
V Bh r h
= ( ) ( ) =
= =
6 35 11 210 11 2
2 315. π ≈ ..
. ft
. , .
14 62 4
452 16 3
62 452 16 28 033
( ) ( )( ) =
( ) ( ) =16. 992
28 033 92 2 000 14 02
45 45 9
, . , .
º º
lb
tons17.
18.
÷ ≈
− − 00º triangle: 2Y 2
using Pythagorean theorem:
a
19.
22 + ( ) =
+ ( ) ( ) =
+ =
=
=
32 2
2 3 3 2
2 9 2 2
10 2 2
10 2
a H
a a a H
a a H
a H
a H == =
+ = ( )+ = ( )( )+ =
a a
L B B
L B B B
L B
2 10 10
2 2 102
2 2 10 10
2 2
20.
BB
L B B
L B
L B B B
2 100
2 2 10 2
2 9 2
9 2 9 2 3
+ =
=
= = =
Systematic Review 21E1.
2.
3.
3
31
1 2 2
16 7 32
16 21
=
( ) =
=22
8 211
8 21
16 72
8 71
8 7
8 2
8
7 2 2 7 4 7 2 14
= =
= =
= = ( ) =
4.
5.
6.
7.
88.
9.
10
7 2
3 10 7 10 3 7 10 10
21 100 21 10 210
( )( ) = ( ) ( ) =
= ( ) =
..
11.
5 7 4 3 5 7 4 3
10 18
2 6
10
+ = +
=
:
cannot be simplified
332
5 31
5 3
112 2 222
121 2 4842 363
363 12
= =
+ =
+ =
=
= =
12. L
L
L
L 11 3 11 3
12
12
11 11 3
60 5 3 2
=
= = ( )( ) =
.
units
a bh
units
13.
114. one triangle:
six
a bh
in
= = ( )( ) =12
12
5 11 14
35 11 2
triangles:
a in
V Bh r h
= ( ) ( ) =
= =
6 35 11 210 11 2
2 315. π ≈ ..
. ft
. , .
14 62 4
452 16 3
62 452 16 28 033
( ) ( )( ) =
( ) ( ) =16. 992
28 033 92 2 000 14 02
45 45 9
, . , .
º º
lb
tons17.
18.
÷ ≈
− − 00º triangle: 2Y 2
using Pythagorean theorem:
a
19.
22 + ( ) =
+ ( ) ( ) =
+ =
=
=
32 2
2 3 3 2
2 9 2 2
10 2 2
10 2
a H
a a a H
a a H
a H
a H == =
+ = ( )+ = ( )( )+ =
a a
L B B
L B B B
L B
2 10 10
2 2 102
2 2 10 10
2 2
20.
BB
L B B
L B
L B B B
2 100
2 2 10 2
2 9 2
9 2 9 2 3
+ =
=
= = =
Lesson Practice 22ALesson Practice 22A1.
2.
axiom; postulate
theorems
33.
4.
5.
6.
7.
converses
congruent
bisector
congruent
coongruent
180º
parallel
360º
if alternate
8.
9.
10.
11. interior angles are
congruent, the two lines ccut
by a transversal are parallel.
if a quad12. rrilateral has two pairs
of parallel sides, it iis
a parallelogram.
in a triangle, if the le13. gg squared
plus the leg squared equals the
hypottenuse squared, it is a
right triangle.
if t14. wwo perpendicular lines
intersect, they form rigght angles.
a quadrilateral with only one pa15. iir
parallel sides is a trapezoid.
property
of
16. of symmetry
reflexive property
transitive
17.
18. property
Lesson Practice 22a - sYsteMatic reVieW 22c
soLUtions GeoMetrY180
Lesson Practice 22A1.
2.
axiom; postulate
theorems
33.
4.
5.
6.
7.
converses
congruent
bisector
congruent
coongruent
180º
parallel
360º
if alternate
8.
9.
10.
11. interior angles are
congruent, the two lines ccut
by a transversal are parallel.
if a quad12. rrilateral has two pairs
of parallel sides, it iis
a parallelogram.
in a triangle, if the le13. gg squared
plus the leg squared equals the
hypottenuse squared, it is a
right triangle.
if t14. wwo perpendicular lines
intersect, they form rigght angles.
a quadrilateral with only one pa15. iir
parallel sides is a trapezoid.
property
of
16. of symmetry
reflexive property
transitive
17.
18. property
Lesson Practice 22BLesson Practice 22B1.
2.
3.
4
assumes
proven
converse
..
5.
6.
isosceles
transversal; congruent
complementaary
exterior
midpoint
or perpedicular bisect
7.
8.
oor
rectangle
square
the measures of sup
( )9.
10.
11. pplementary
add up to 180º.
a quadrila
angles
12. tteral with two pairs of
sides is a parparallel aallelogram.
if alternate exterior angles are13.
congruent, the lines cut by a
transversal are parallel.
if two line segments are congruen14. tt, they
are equal in length.
a polygon with 15. ttwo pairs of
parallel sides and four congruent
sides is a rhombus.
transitive property
p
16.
17. rroperty of symmetry
reflexive property18.
Lesson Practice 22B1.
2.
3.
4
assumes
proven
converse
..
5.
6.
isosceles
transversal; congruent
complementaary
exterior
midpoint
or perpedicular bisect
7.
8.
oor
rectangle
square
the measures of sup
( )9.
10.
11. pplementary
add up to 180º.
a quadrila
angles
12. tteral with two pairs of
sides is a parparallel aallelogram.
if alternate exterior angles are13.
congruent, the lines cut by a
transversal are parallel.
if two line segments are congruen14. tt, they
are equal in length.
a polygon with 15. ttwo pairs of
parallel sides and four congruent
sides is a rhombus.
transitive property
p
16.
17. rroperty of symmetry
reflexive property18.
Systematic Review 22CSystematic Review 22C1.
2.
3.
unproven
congruent
corrresponding, alternate interior,
and exterior anngles are
congruent.
interior angles
4.
5.
6.
a B c+ >
iif a B then B a= =
( ) =
= = =
( ) =
7.
8.
5 6 2 10 6
5 6 3 5 18 5 9 2
5 3 2 115 2
13
2
13 2
2 2
13 2
4
13 22
13 22
2 3
5
2 3 5
5 5
2 1
9.
10.
11.
= = =
= = 55
25
2 155
2 132 172
2 169 2892 120
120 4 30
=
+ =
+ =
=
= =
12. L
L
L
L ==
= − =
=
2 30
180 120 60
360 60 6
13. exterior angle º º º
º º÷ ssides
r
in
14. circle: c
sec
= ( ) ( ) ( )=
2 2 3 14 10
62 8
π ≈ .
.
ttor is 150º360º
circle:=
( )
512
512
62 8 26 17. .
of
≈ iin
r
in
15. circle: a
sector i
= ( ) ( )=
π ≈2 3 14 102
314 2
.
ss 150º360º
circle:=
( )
512
512
314 130 83 2.
of
in≈
16.. one triangle:
eight tri
a bh= = ( ) ( ) =12
12
3 13 10 15 13
aangles:
a = ( ) ( ) =
= =
8 15 13 120 13 432 67 2
2
≈
π
. in
V Bh r17. hh ≈ 3 14 7 52 4
706 5 3
62 706 5 4
. .
. ft
.
( ) ( )( )
=
( ) ( ) =18. 33 803
43 803 2 000 21 9
13
,
, , .
lb
tons19.
20.
÷ ≈
cone: V = BBh r h=
( ) ( ) ( ) =
13
2
13
3 14 142 16 5 3 384 92 3
π ≈
. . , . ft
cylinnder:
t
V Bh r h= =
( ) ( )( ) =
π ≈2
3 14 142 17 10 462 48 3. , . ft
ootal: 3 384 92 10 462 48
13 847 4 3
, . , .
, . ft
+ =
sYsteMatic reVieW 22c - sYsteMatic reVieW 22e
soLUtionsGeoMetrY 181
Systematic Review 22C1.
2.
3.
unproven
congruent
corrresponding, alternate interior,
and exterior anngles are
congruent.
interior angles
4.
5.
6.
a B c+ >
iif a B then B a= =
( ) =
= = =
( ) =
7.
8.
5 6 2 10 6
5 6 3 5 18 5 9 2
5 3 2 115 2
13
2
13 2
2 2
13 2
4
13 22
13 22
2 3
5
2 3 5
5 5
2 1
9.
10.
11.
= = =
= = 55
25
2 155
2 132 172
2 169 2892 120
120 4 30
=
+ =
+ =
=
= =
12. L
L
L
L ==
= − =
=
2 30
180 120 60
360 60 6
13. exterior angle º º º
º º÷ ssides
r
in
14. circle: c
sec
= ( ) ( ) ( )=
2 2 3 14 10
62 8
π ≈ .
.
ttor is 150º360º
circle:=
( )
512
512
62 8 26 17. .
of
≈ iin
r
in
15. circle: a
sector i
= ( ) ( )=
π ≈2 3 14 102
314 2
.
ss 150º360º
circle:=
( )
512
512
314 130 83 2.
of
in≈
16.. one triangle:
eight tri
a bh= = ( ) ( ) =12
12
3 13 10 15 13
aangles:
a = ( ) ( ) =
= =
8 15 13 120 13 432 67 2
2
≈
π
. in
V Bh r17. hh ≈ 3 14 7 52 4
706 5 3
62 706 5 4
. .
. ft
.
( ) ( )( )
=
( ) ( ) =18. 33 803
43 803 2 000 21 9
13
,
, , .
lb
tons19.
20.
÷ ≈
cone: V = BBh r h=
( ) ( ) ( ) =
13
2
13
3 14 142 16 5 3 384 92 3
π ≈
. . , . ft
cylinnder:
t
V Bh r h= =
( ) ( )( ) =
π ≈2
3 14 142 17 10 462 48 3. , . ft
ootal: 3 384 92 10 462 48
13 847 4 3
, . , .
, . ft
+ =
Systematic Review 22DSystematic Review 22D1.
2.
3.
proven
congruent
isosceeles
square4.
5.
6.
7.
8.
9.
10.
11
360
5 2 10
5 3
9 2
9
º
a a=
( ) =
..
12.
H
H
H
H
H units
a
2 122 162
2 144 2562 400
400
20
= +
= +
=
=
=
== = ( ) ( ) =
=
12
12
12 16 96 2
18
bh units
13. exterior angle 00 150 30
360 30 12
º º º
º º
− =
=÷ sides
14.
15.
see drawing
onne triangle:
ten triang
a bh= = ( ) ( ) =12
12
6 13 25 75 13
lles:
a
in
V Bh r h
= ( ) =
= =
10 75 13 750 13
2 704 16 2
2
≈
π ≈
, .
16. 33 14 92 4
1 017 36 3
62 1 017 36 63
.
, . ft
, .
( ) ( )( ) =
( ) ( ) =17. ,, .
, . , .
076 32
63 076 32 2 000 31 54
lb
tons18.
19.
÷ ≈
"rooff":
triangles:
a
a m
bh
= ( ) ( ) =
= ( ) = (
2 12 6 7 160 8 2
2 12
2
. .
)) ( ) ( ) =
= ( ) ( ) + ( ) ( ) =
12
6 6 36 2
2 6 5 2 12 5 180 2
m
m
sides:
a
bbottom:
a
total:
a
= ( ) ( ) =
= + + + =
6 12 72 2
160 8 36 180 72.
m
4448 8 2
12
6 6 12
216 3
. m
V Bh
m
20. prism:
rect
= = ( ) ( ) ( ) =
aangular solid:
V
total: V
= ( ) ( ) ( ) =
= +
6 12 5 360 3
216
m
3360 576 3= m
Systematic Review 22D1.
2.
3.
proven
congruent
isosceeles
square4.
5.
6.
7.
8.
9.
10.
11
360
5 2 10
5 3
9 2
9
º
a a=
( ) =
..
12.
H
H
H
H
H units
a
2 122 162
2 144 2562 400
400
20
= +
= +
=
=
=
== = ( ) ( ) =
=
12
12
12 16 96 2
18
bh units
13. exterior angle 00 150 30
360 30 12
º º º
º º
− =
=÷ sides
14.
15.
see drawing
onne triangle:
ten triang
a bh= = ( ) ( ) =12
12
6 13 25 75 13
lles:
a
in
V Bh r h
= ( ) =
= =
10 75 13 750 13
2 704 16 2
2
≈
π ≈
, .
16. 33 14 92 4
1 017 36 3
62 1 017 36 63
.
, . ft
, .
( ) ( )( ) =
( ) ( ) =17. ,, .
, . , .
076 32
63 076 32 2 000 31 54
lb
tons18.
19.
÷ ≈
"rooff":
triangles:
a
a m
bh
= ( ) ( ) =
= ( ) = (
2 12 6 7 160 8 2
2 12
2
. .
)) ( ) ( ) =
= ( ) ( ) + ( ) ( ) =
12
6 6 36 2
2 6 5 2 12 5 180 2
m
m
sides:
a
bbottom:
a
total:
a
= ( ) ( ) =
= + + + =
6 12 72 2
160 8 36 180 72.
m
4448 8 2
12
6 6 12
216 3
. m
V Bh
m
20. prism:
rect
= = ( ) ( ) ( ) =
aangular solid:
V
total: V
= ( ) ( ) ( ) =
= +
6 12 5 360 3
216
m
3360 576 3= m
Systematic Review 22D1.
2.
3.
proven
congruent
isosceeles
square4.
5.
6.
7.
8.
9.
10.
11
360
5 2 10
5 3
9 2
9
º
a a=
( ) =
..
12.
H
H
H
H
H units
a
2 122 162
2 144 2562 400
400
20
= +
= +
=
=
=
== = ( ) ( ) =
=
12
12
12 16 96 2
18
bh units
13. exterior angle 00 150 30
360 30 12
º º º
º º
− =
=÷ sides
14.
15.
see drawing
onne triangle:
ten triang
a bh= = ( ) ( ) =12
12
6 13 25 75 13
lles:
a
in
V Bh r h
= ( ) =
= =
10 75 13 750 13
2 704 16 2
2
≈
π ≈
, .
16. 33 14 92 4
1 017 36 3
62 1 017 36 63
.
, . ft
, .
( ) ( )( ) =
( ) ( ) =17. ,, .
, . , .
076 32
63 076 32 2 000 31 54
lb
tons18.
19.
÷ ≈
"rooff":
triangles:
a
a m
bh
= ( ) ( ) =
= ( ) = (
2 12 6 7 160 8 2
2 12
2
. .
)) ( ) ( ) =
= ( ) ( ) + ( ) ( ) =
12
6 6 36 2
2 6 5 2 12 5 180 2
m
m
sides:
a
bbottom:
a
total:
a
= ( ) ( ) =
= + + + =
6 12 72 2
160 8 36 180 72.
m
4448 8 2
12
6 6 12
216 3
. m
V Bh
m
20. prism:
rect
= = ( ) ( ) ( ) =
aangular solid:
V
total: V
= ( ) ( ) ( ) =
= +
6 12 5 360 3
216
m
3360 576 3= m
Systematic Review 22ESystematic Review 22E1.
2.
3.
rectangle
right angle
rrhombus
supplementary
equal measures
4.
5.
6. if a B= and B c then a c= =
=
=
7.
8.
9.
10.
11.
162
8
8 3
9 2
9
6
4 8
6 2
4 88 2
12
4 16
4 34 4
2 316
38
482 2 502
2 304 2 2
= = ( ) =
=
+ =
+ =
12. L
L, ,,
.
500
2 196
196 14
2 2 3
L
L units
r
=
= =
= ( )13. circle:
C π ≈ 114 6 37 68( ) ( ) =
=
. in
whole circlsector is 360º360º
ee
c in
a r
= ( ) =
= ( ) (
1 37 68 37 68
2 3 14 62
. .
.
14. circle:
π ≈ )) =
( ) =
113 04 2
1 113 04 113 04 2
.
. .
in
insector:
one15. triangle:
five tri
a bh in= = ( )( ) =12
12
8 10 22 88 10 2
aangles:
a = ( ) ( ) =
= =
5 88 10 440 10
1 391 40 2
≈
, . in
V Bh16. ≠≠r h2 3 14 122 4
1 808 64 3
62 1 808 6
⊕( ) ( )( ) =
( )
.
, . ft
, .17. 44 112 135 68
112 135 68 2 000 56 07
( ) = , .
, . , .
lb
tons÷ ≈
18..
19.
X
X X
X X X
X X
X
4 81
2 9 2 9
3 3 2 9
3 9
− =
−( ) +( ) =
−( ) +( ) +( )− =
(( ) −( ) =
( ) −( ) +( )
− =
( ) −( ) =
X
X X X
X X
X X
X
2 9
3 3
4 25 2
2 2 25
20.
22 5 5( ) −( ) +( )X X
sYsteMatic reVieW 22e - Lesson Practice 23B
soLUtions GeoMetrY182
Systematic Review 22E1.
2.
3.
rectangle
right angle
rrhombus
supplementary
equal measures
4.
5.
6. if a B= and B c then a c= =
=
=
7.
8.
9.
10.
11.
162
8
8 3
9 2
9
6
4 8
6 2
4 88 2
12
4 16
4 34 4
2 316
38
482 2 502
2 304 2 2
= = ( ) =
=
+ =
+ =
12. L
L, ,,
.
500
2 196
196 14
2 2 3
L
L units
r
=
= =
= ( )13. circle:
C π ≈ 114 6 37 68( ) ( ) =
=
. in
whole circlsector is 360º360º
ee
c in
a r
= ( ) =
= ( ) (
1 37 68 37 68
2 3 14 62
. .
.
14. circle:
π ≈ )) =
( ) =
113 04 2
1 113 04 113 04 2
.
. .
in
insector:
one15. triangle:
five tri
a bh in= = ( )( ) =12
12
8 10 22 88 10 2
aangles:
a = ( ) ( ) =
= =
5 88 10 440 10
1 391 40 2
≈
, . in
V Bh16. ≠≠r h2 3 14 122 4
1 808 64 3
62 1 808 6
⊕( ) ( )( ) =
( )
.
, . ft
, .17. 44 112 135 68
112 135 68 2 000 56 07
( ) = , .
, . , .
lb
tons÷ ≈
18..
19.
X
X X
X X X
X X
X
4 81
2 9 2 9
3 3 2 9
3 9
− =
−( ) +( ) =
−( ) +( ) +( )− =
(( ) −( ) =
( ) −( ) +( )
− =
( ) −( ) =
X
X X X
X X
X X
X
2 9
3 3
4 25 2
2 2 25
20.
22 5 5( ) −( ) +( )X X
Systematic Review 22E1.
2.
3.
rectangle
right angle
rrhombus
supplementary
equal measures
4.
5.
6. if a B= and B c then a c= =
=
=
7.
8.
9.
10.
11.
162
8
8 3
9 2
9
6
4 8
6 2
4 88 2
12
4 16
4 34 4
2 316
38
482 2 502
2 304 2 2
= = ( ) =
=
+ =
+ =
12. L
L, ,,
.
500
2 196
196 14
2 2 3
L
L units
r
=
= =
= ( )13. circle:
C π ≈ 114 6 37 68( ) ( ) =
=
. in
whole circlsector is 360º360º
ee
c in
a r
= ( ) =
= ( ) (
1 37 68 37 68
2 3 14 62
. .
.
14. circle:
π ≈ )) =
( ) =
113 04 2
1 113 04 113 04 2
.
. .
in
insector:
one15. triangle:
five tri
a bh in= = ( )( ) =12
12
8 10 22 88 10 2
aangles:
a = ( ) ( ) =
= =
5 88 10 440 10
1 391 40 2
≈
, . in
V Bh16. ≠≠r h2 3 14 122 4
1 808 64 3
62 1 808 6
⊕( ) ( )( ) =
( )
.
, . ft
, .17. 44 112 135 68
112 135 68 2 000 56 07
( ) = , .
, . , .
lb
tons÷ ≈
18..
19.
X
X X
X X X
X X
X
4 81
2 9 2 9
3 3 2 9
3 9
− =
−( ) +( ) =
−( ) +( ) +( )− =
(( ) −( ) =
( ) −( ) +( )
− =
( ) −( ) =
X
X X X
X X
X X
X
2 9
3 3
4 25 2
2 2 25
20.
22 5 5( ) −( ) +( )X X
Lesson Practice 23ALesson Practice 23A1.
2.
3.
4.
5.
6.
∠
∠
D
DFe
DeF
a
ac
DF
77.
8.
9.
10.
11.
12.
13.
∠
∠
XtY
rst
Yt
tr
YtX
Xt
correspond
114.
15.
16.
1
exterior
remote interior
180 120 60º º º− =
77.
18.
180 89 91
60 91 151
º º º
º º º
− =
∠ = ∠ + ∠
+ =
m D m B m c
remote interior angles( )
Lesson Practice 23BLesson Practice 23B1.
2.
3.
4.
5.
6.
BD
BF
BDF
c
aec
ac
∠
∠
77.
8.
9.
10.
11.
12.
13.
14.
Bc
ce
Bca
B
e
De
∠
∠
180
exterior
115.
16.
17.
18.
B c
m D m
,
º º º
º º º
180 84 96
180 132 48
− =
− =
∠ = ∠∠ + ∠
+ =
( )
B m c
96 48 144º º º
remote interior angles
Lesson Practice 23B - sYsteMatic reVieW 23D
soLUtionsGeoMetrY 183
Lesson Practice 23B1.
2.
3.
4.
5.
6.
BD
BF
BDF
c
aec
ac
∠
∠
77.
8.
9.
10.
11.
12.
13.
14.
Bc
ce
Bca
B
e
De
∠
∠
180
exterior
115.
16.
17.
18.
B c
m D m
,
º º º
º º º
180 84 96
180 132 48
− =
− =
∠ = ∠∠ + ∠
+ =
( )
B m c
96 48 144º º º
remote interior angles
Systematic Review 23CSystematic Review 23C1.
2.
3.
4.
5.
LM
JG
JHG
MKL
LMK
∠
∠
66.
7.
8.
180 123 57
180 110 70
180
º º º
º º º
º
− =
− =
∠ = − ∠ +m a m B m∠∠( )∠ = − +( )∠ = − =
−
c
m a
m a
180 57 70
180 127 53
30 6
º º º
º º º
º9. 00 90
2
5 2 10
º º−
= ×
× =
triangle
hypotenuse short leg
10. 330 60 90
3
5 3 5 3
º º º− −
= ×
× =
triangle
long leg short leg
111.
12.
13.
parallel sides
regular
complementary anggles
parallelogram
leg squared; leg square
14.
15. dd
exterior angles of a polygon
add up to 360º
16.
.. if each has
a measure of 120º, there must
be 3660º 120, or 3 sides.
triangle
÷
π ≈17. V r= 43
3 43
3 14.(( ) ( ) =
( ) ( )
=
4 33
43
3 14 79 507 332 87 3
4
.
. . .≈
π
cm
sa r18. 22 4 3 14 4 32
4 3 14 18 49 232 23 2
≈
≈
. .
. . .
( ) ( ) =
( ) ( ) cm
ar19. eea of circle with r
a r .
.
=
= ( ) ( ) =
( )
4:
π ≈2 3 14 42
3 14 166 50 24 2
45
( ) = . ft
º is 18
of 360º,
so area of arc ==
=
=
=
18
of 50.24 ft2
(area
50 24 8 6 28 2. . ft÷
V Bh oof base times height)
V = ( )6 28 5 2 32 66 3
3
. . . ft≈
20. 22 66 3 62 3
2 024 92
. ft / ft
, .
lb
lb
( ) =
using rounded annswer from #19( )
Systematic Review 23C1.
2.
3.
4.
5.
LM
JG
JHG
MKL
LMK
∠
∠
66.
7.
8.
180 123 57
180 110 70
180
º º º
º º º
º
− =
− =
∠ = − ∠ +m a m B m∠∠( )∠ = − +( )∠ = − =
−
c
m a
m a
180 57 70
180 127 53
30 6
º º º
º º º
º9. 00 90
2
5 2 10
º º−
= ×
× =
triangle
hypotenuse short leg
10. 330 60 90
3
5 3 5 3
º º º− −
= ×
× =
triangle
long leg short leg
111.
12.
13.
parallel sides
regular
complementary anggles
parallelogram
leg squared; leg square
14.
15. dd
exterior angles of a polygon
add up to 360º
16.
.. if each has
a measure of 120º, there must
be 3660º 120, or 3 sides.
triangle
÷
π ≈17. V r= 43
3 43
3 14.(( ) ( ) =
( ) ( )
=
4 33
43
3 14 79 507 332 87 3
4
.
. . .≈
π
cm
sa r18. 22 4 3 14 4 32
4 3 14 18 49 232 23 2
≈
≈
. .
. . .
( ) ( ) =
( ) ( ) cm
ar19. eea of circle with r
a r .
.
=
= ( ) ( ) =
( )
4:
π ≈2 3 14 42
3 14 166 50 24 2
45
( ) = . ft
º is 18
of 360º,
so area of arc ==
=
=
=
18
of 50.24 ft2
(area
50 24 8 6 28 2. . ft÷
V Bh oof base times height)
V = ( )6 28 5 2 32 66 3
3
. . . ft≈
20. 22 66 3 62 3
2 024 92
. ft / ft
, .
lb
lb
( ) =
using rounded annswer from #19( )
Systematic Review 23DSystematic Review 23D1.
2.
3.
4.
5.
Za
YX
XaZ
ZaX
aXZ
∠
66.
7.
m c
m c
m X
∠ = − +( )∠ = − =
∠ =
180 52 59
180 111 69
180
º º º
º º º
ºº
º º º
º
− ∠
∠ = − =
∠ = ∠ +
m c
m X
m Y m c
180 69 111
528.
remote interrior angles( )∠ = + =
+ =
+
m Y
a
a
69 52 121
2 482 502
2 2
º º º
,
9.
3304 2 500
2 19614
102 112 2
100 121
=
==
+ =
+ =
,
aa
B
units
10.
BB
B
B
quadrilateral or
2
221 2
221
=
= units
paralle11. llogram
perpendicular
congruent
180º
co
12.
13.
14.
15. nngruent
exterior angles add up to
360º;
16.
360 6÷ 00 6
2 2
1
=
= × ( )
;
min
sides
hexagon
a major or17. π ≈
002
152
3 14
5 7 5 3 14 117 75 2
× ( ) =
( ) ( ) ( ) =
.
. . . m
18. cchange 42 inches to feet:
42 12 3 5
13
13
÷ =
= =
. ft
V Bh ππ ≈
≈
r h2
13
3 14 3 52
15
13
3 14 12 25 15 19
. .
. .
( ) ( ) ( ) =
( ) ( ) ( ) 22 33 3
60 16
360
. ft
º º19. = of so area of base
will bbe 16
that of the whole circle.
a r= ( ) ( )π ≈2 3 14 7.22
3 14 49 153 86 2
153 86 6 25 643 2
=
( ) ( ) =
=
. . ft
. . ft÷ ≈
V Bh == ×
( ) =
25 643 2 51 29 3
51 29 62 3 3 1
. . ft
. / ft ,
≈
20. ft3 lb 779 98. lb
using rounded answer from #19( )
sYsteMatic reVieW 23D - sYsteMatic reVieW 23e
soLUtions GeoMetrY184
Systematic Review 23D1.
2.
3.
4.
5.
Za
YX
XaZ
ZaX
aXZ
∠
66.
7.
m c
m c
m X
∠ = − +( )∠ = − =
∠ =
180 52 59
180 111 69
180
º º º
º º º
ºº
º º º
º
− ∠
∠ = − =
∠ = ∠ +
m c
m X
m Y m c
180 69 111
528.
remote interrior angles( )∠ = + =
+ =
+
m Y
a
a
69 52 121
2 482 502
2 2
º º º
,
9.
3304 2 500
2 19614
102 112 2
100 121
=
==
+ =
+ =
,
aa
B
units
10.
BB
B
B
quadrilateral or
2
221 2
221
=
= units
paralle11. llogram
perpendicular
congruent
180º
co
12.
13.
14.
15. nngruent
exterior angles add up to
360º;
16.
360 6÷ 00 6
2 2
1
=
= × ( )
;
min
sides
hexagon
a major or17. π ≈
002
152
3 14
5 7 5 3 14 117 75 2
× ( ) =
( ) ( ) ( ) =
.
. . . m
18. cchange 42 inches to feet:
42 12 3 5
13
13
÷ =
= =
. ft
V Bh ππ ≈
≈
r h2
13
3 14 3 52
15
13
3 14 12 25 15 19
. .
. .
( ) ( ) ( ) =
( ) ( ) ( ) 22 33 3
60 16
360
. ft
º º19. = of so area of base
will bbe 16
that of the whole circle.
a r= ( ) ( )π ≈2 3 14 7.22
3 14 49 153 86 2
153 86 6 25 643 2
=
( ) ( ) =
=
. . ft
. . ft÷ ≈
V Bh == ×
( ) =
25 643 2 51 29 3
51 29 62 3 3 1
. . ft
. / ft ,
≈
20. ft3 lb 779 98. lb
using rounded answer from #19( )
Systematic Review 23ESystematic Review 23E1.
2.
3.
4.
5.
Vt
QtV
tVQ
tsW
s
∠
∠
WWt
m a
m X
6.
7.
∠ = − +( ) =
− ( ) =
∠ =
180 62 67
180 129 51
18
º º º
º º º
00 67 113
180 62 118
2 342 362
2 1
º º º
º º º
− =
∠ = − =
+ =
+
8.
9.
m Y
a
a ,, ,156 1 296
2 140
140
4 35 2 35
102 102 2
100
=
=
=
= =
+ =
a
a
a
B10.
++ =
=
=
= =
100 2
200 2
200
100 2 10 2
12
B
B
B
B
vertex11.
12.
13. allternate; congruent
the sum of the
14.
15.
180º
;> lengths of
the two shorter sides of a
triangle must be greater than
the length of the long sidde.
ctogon has 8 sides. each
exterior angle
16. o
mmust have a
measure of 360º so
i
÷8 45, º,or
each nnterior angle must have
a measure of 180º − =45 1º 335
33 12 3 96
13
2
13
3 14 3
º.
. ft .
.
17. × =
= =
( )
in
V Bh r hπ ≈
.. .962
7 114 89 3
2 6 8 0
4 2 0
( ) ( )
+ + =
+( ) +( ) =
+
≈ in
X X
X X
X
18.
44 04
2 02
2 5 62 5 6 0
3 2
== −
+ == −
+ = −
+ + =
+( ) +( )
X
XX
X X
X X
X X
19.
==
+ == −
+ == −
− + =
− − =
−
0
3 03
2 02
2 3 2 302 3 28 0
7
XX
XX
X X
X X
X
20.
(( ) +( ) =
− =
=
+ == −
X
X
X
XX
4 0
7 0
7
4 04
Systematic Review 23E1.
2.
3.
4.
5.
Vt
QtV
tVQ
tsW
s
∠
∠
WWt
m a
m X
6.
7.
∠ = − +( ) =
− ( ) =
∠ =
180 62 67
180 129 51
18
º º º
º º º
00 67 113
180 62 118
2 342 362
2 1
º º º
º º º
− =
∠ = − =
+ =
+
8.
9.
m Y
a
a ,, ,156 1 296
2 140
140
4 35 2 35
102 102 2
100
=
=
=
= =
+ =
a
a
a
B10.
++ =
=
=
= =
100 2
200 2
200
100 2 10 2
12
B
B
B
B
vertex11.
12.
13. allternate; congruent
the sum of the
14.
15.
180º
;> lengths of
the two shorter sides of a
triangle must be greater than
the length of the long sidde.
ctogon has 8 sides. each
exterior angle
16. o
mmust have a
measure of 360º so
i
÷8 45, º,or
each nnterior angle must have
a measure of 180º − =45 1º 335
33 12 3 96
13
2
13
3 14 3
º.
. ft .
.
17. × =
= =
( )
in
V Bh r hπ ≈
.. .962
7 114 89 3
2 6 8 0
4 2 0
( ) ( )
+ + =
+( ) +( ) =
+
≈ in
X X
X X
X
18.
44 04
2 02
2 5 62 5 6 0
3 2
== −
+ == −
+ = −
+ + =
+( ) +( )
X
XX
X X
X X
X X
19.
==
+ == −
+ == −
− + =
− − =
−
0
3 03
2 02
2 3 2 302 3 28 0
7
XX
XX
X X
X X
X
20.
(( ) +( ) =
− =
=
+ == −
X
X
X
XX
4 0
7 0
7
4 04
sYsteMatic reVieW 23e - sYsteMatic reVieW 24c
soLUtionsGeoMetrY 185
Systematic Review 23E1.
2.
3.
4.
5.
Vt
QtV
tVQ
tsW
s
∠
∠
WWt
m a
m X
6.
7.
∠ = − +( ) =
− ( ) =
∠ =
180 62 67
180 129 51
18
º º º
º º º
00 67 113
180 62 118
2 342 362
2 1
º º º
º º º
− =
∠ = − =
+ =
+
8.
9.
m Y
a
a ,, ,156 1 296
2 140
140
4 35 2 35
102 102 2
100
=
=
=
= =
+ =
a
a
a
B10.
++ =
=
=
= =
100 2
200 2
200
100 2 10 2
12
B
B
B
B
vertex11.
12.
13. allternate; congruent
the sum of the
14.
15.
180º
;> lengths of
the two shorter sides of a
triangle must be greater than
the length of the long sidde.
ctogon has 8 sides. each
exterior angle
16. o
mmust have a
measure of 360º so
i
÷8 45, º,or
each nnterior angle must have
a measure of 180º − =45 1º 335
33 12 3 96
13
2
13
3 14 3
º.
. ft .
.
17. × =
= =
( )
in
V Bh r hπ ≈
.. .962
7 114 89 3
2 6 8 0
4 2 0
( ) ( )
+ + =
+( ) +( ) =
+
≈ in
X X
X X
X
18.
44 04
2 02
2 5 62 5 6 0
3 2
== −
+ == −
+ = −
+ + =
+( ) +( )
X
XX
X X
X X
X X
19.
==
+ == −
+ == −
− + =
− − =
−
0
3 03
2 02
2 3 2 302 3 28 0
7
XX
XX
X X
X X
X
20.
(( ) +( ) =
− =
=
+ == −
X
X
X
XX
4 0
7 0
7
4 04
Lesson Practice 24ALesson Practice 24APlease note: in some cases, tthe student
will be able to do the proof in a sslightly
different way from the way given in
thee answer key. order of statements is
not importtant except where one
statement depends upon annother, in
which case the supporting statement
wwill need to be made first.
given
definition
1.
2. of a square
reflexive property
sss postulat
3.
4. ee
given
given
transversal; congruent
sas p
5.
6.
7.
8. oostulate
order of 9 and 10
may be rev
9. aB cB;
eersed
given
10.
11.
12.
13.
aD cD
BD BD
aBD cBD
;
;
;
114.
15.
16.
given
is a rhombus
aD order of
aBcD
cB≅ ( 16 and 17
may be reversed)
17.
18.
19.
aB cD
BD BD
≅
≅
aBD cDB≅
Lesson Practice 24BLesson Practice 24B1.
2.
3.
4.
given
given
defin
BD BD≅
iition of bisector
given
definition of a rect
5.
6. aangle
interior
reflexive pr
7.
8.
alternate angles
ooperty
given
definition of perpendicu
9.
10.
11.
sas
llar
right angles are congruent
given
def
12.
13.
14. iinition of midpoint
reflexive property
sas
15.
16. postulate
given
given
(19 and 20
17.
18.
19. ec Bc≅ mmay be reversed)
ac20.
21.
≅
≅
Dc
aBc Dec
Systematic Review 24CSystematic Review 24C1.
2.
definition of square
deffinition of square
reflexive property
sQU
3.
4. ≅ UUrs
5. yes: the diagonal of the square
is a transsversal across the parallel
sides of the squaree, making the
angles named alternate
interior anngles.
the l
6.
7.
8.
9.
10.
11.
12.
126
65
54
61
º
º
º
º
GFe
Fe
eength of the hypotenuse
of a 45º trian− −45 90º º ggle is
equal to 2 times a leg
of the triangle.
33 2
3 2 legs of a 45º
2 3 4 3 2 6
45 9
( ) = ( ) = ( ) =
− −13. ºthe 00º
triangle are congruent
to one another.
t14. hhe length of the hypotenuse
of a 30º t− −60 90º º rriangle is
equal to 2 times the short leg
of thhe triangle.
2
the length of the long
3 2 4 3( ) =
15. lleg of a
30º triangle is equal to
3 t
− −60 90º º
iimes the short leg of the triangle.
2 3 3 2 9( ) = ( ) = 22 3 6
180
2
( ) =
16.
17.
18.
19.
trapezoid
º
Volume of cylinnder
V Bh r h
:
.
. ft
= = ( ) ( ) ( ) =π ≈2 3 14 42
8
401 92 3
section is 90º360º
of the
cylinder
14
=
× =
14
401 92 100 48. . ft
. ftft
, .
3
100 48 3 623
6 229 7620. x lblb =
sYsteMatic reVieW 24c - sYsteMatic reVieW 24e
soLUtions GeoMetrY186
Systematic Review 24C1.
2.
definition of square
deffinition of square
reflexive property
sQU
3.
4. ≅ UUrs
5. yes: the diagonal of the square
is a transsversal across the parallel
sides of the squaree, making the
angles named alternate
interior anngles.
the l
6.
7.
8.
9.
10.
11.
12.
126
65
54
61
º
º
º
º
GFe
Fe
eength of the hypotenuse
of a 45º trian− −45 90º º ggle is
equal to 2 times a leg
of the triangle.
33 2
3 2 legs of a 45º
2 3 4 3 2 6
45 9
( ) = ( ) = ( ) =
− −13. ºthe 00º
triangle are congruent
to one another.
t14. hhe length of the hypotenuse
of a 30º t− −60 90º º rriangle is
equal to 2 times the short leg
of thhe triangle.
2
the length of the long
3 2 4 3( ) =
15. lleg of a
30º triangle is equal to
3 t
− −60 90º º
iimes the short leg of the triangle.
2 3 3 2 9( ) = ( ) = 22 3 6
180
2
( ) =
16.
17.
18.
19.
trapezoid
º
Volume of cylinnder
V Bh r h
:
.
. ft
= = ( ) ( ) ( ) =π ≈2 3 14 42
8
401 92 3
section is 90º360º
of the
cylinder
14
=
× =
14
401 92 100 48. . ft
. ftft
, .
3
100 48 3 623
6 229 7620. x lblb =
Systematic Review 24DSystematic Review 24D1. definition of isosceles ttriangle
definition of bisector
reflexive pr
2.
3. ooperty
cDM
yes; corresponding angles
a
4.
5.
≅ cFM
rre congruent
69º
divide hyp
6.
7.
8.
9.
10.
111º
∠DGe
Ge
ootenuse by 2:
4
2
same as a, s
= = = =4 2
2 2
4 2
4
4 22
2 2
11. oo 2 2
First, find hy
12.
13.
a bh in= = ( ) ( ) =12
12
4 9 18 2
ppotenuse:
92
sum of
+ =
+ =
=
=
=
42 2
81 16 2
97 2
97 9 8
H
H
H
H
P
≈ .
sides =
+ + =
+ =
4 9 9 8 22 8
2 2 2
. . in
leg leg hypotenuse14.
or a B c2 2 2+ =
15.
16.
17.
postulates
congruent
isoscelles
square
olume of cylinder:
18.
19. V
V Bh r h= = π ≈2 3.114 102
12
3 768 3
13
( ) ( ) ( ) =
=
, ft
section is 120º360º
oof the
cylinder
13
× =3 768 3 1 256 3
1 256
, ft , ft
,20. fftft
,3 623
77 872× = lb lb
Systematic Review 24D1. definition of isosceles ttriangle
definition of bisector
reflexive pr
2.
3. ooperty
cDM
yes; corresponding angles
a
4.
5.
≅ cFM
rre congruent
69º
divide hyp
6.
7.
8.
9.
10.
111º
∠DGe
Ge
ootenuse by 2:
4
2
same as a, s
= = = =4 2
2 2
4 2
4
4 22
2 2
11. oo 2 2
First, find hy
12.
13.
a bh in= = ( ) ( ) =12
12
4 9 18 2
ppotenuse:
92
sum of
+ =
+ =
=
=
=
42 2
81 16 2
97 2
97 9 8
H
H
H
H
P
≈ .
sides =
+ + =
+ =
4 9 9 8 22 8
2 2 2
. . in
leg leg hypotenuse14.
or a B c2 2 2+ =
15.
16.
17.
postulates
congruent
isoscelles
square
olume of cylinder:
18.
19. V
V Bh r h= = π ≈2 3.114 102
12
3 768 3
13
( ) ( ) ( ) =
=
, ft
section is 120º360º
oof the
cylinder
13
× =3 768 3 1 256 3
1 256
, ft , ft
,20. fftft
,3 623
77 872× = lb lb
Systematic Review 24ESystematic Review 24E1. rH sB≅ (1 and 2 may be reeversed.)
sss postulate
yes; c
2.
3.
4.
5.
rB sH
BH BH
≅
≅
oorresponding angles
are congruent
51º6.
7.
8.
113º
∠∠
+ =
+ =
=
=
tWV
Vt
LL
LL
LL
LL
9.
10. 102 2 202
100 2 4002 300
300
LLL
this
= =100 3 10 3
11. triangle's sides follow the
pattern of 30º
triangles, so a
− −
∠ =
60 90
60
º º
º
12.. this triangle's sides follow the
pattern of 300º
triangles, so b
− −
∠ =
= = (
60 90
30
12
12
10
º º
º
13. a bh )) ( ) =
= + + =
10 3 50 3
86 60
20 10 10 3 30
≈ . square units
14. P ++10 3
47 32≈ . units
15. exterior angles must add to 360º.
since there 10, each exterior
angle mustt have a measure
of 360º10
36º. each interioor rr
angle, therefore will have a
measure of 180º −− =36 144º º.
16. arc Bac has a measure of 264º,
so arc Bc has a measure of
360º the me− =264 96º º. aasure
of the inscribed angle is half the
measurre of the arc,
so m Bac∠ = =
+ =
12
96 48
3
5
7
2
3 5
5 5
x º º
17. ++ =
+ = + =
( )( ) +
( )( ) = +
7 2
2 2
3 5
25
7 2
4
3 55
7 22
2 3 52 5
5 7 25 2
6 5 335 210
1 450 000 1 45 106
0076 7 6 10 3
18.
19.
20.
, , .
. .
= ×
= × −
6640 000 000 000 6 4 1011, , , .= ×
sYsteMatic reVieW 24e - Lesson Practice 25B
soLUtionsGeoMetrY 187
Systematic Review 24E1. rH sB≅ (1 and 2 may be reeversed.)
sss postulate
yes; c
2.
3.
4.
5.
rB sH
BH BH
≅
≅
oorresponding angles
are congruent
51º6.
7.
8.
113º
∠∠
+ =
+ =
=
=
tWV
Vt
LL
LL
LL
LL
9.
10. 102 2 202
100 2 4002 300
300
LLL
this
= =100 3 10 3
11. triangle's sides follow the
pattern of 30º
triangles, so a
− −
∠ =
60 90
60
º º
º
12.. this triangle's sides follow the
pattern of 300º
triangles, so b
− −
∠ =
= = (
60 90
30
12
12
10
º º
º
13. a bh )) ( ) =
= + + =
10 3 50 3
86 60
20 10 10 3 30
≈ . square units
14. P ++10 3
47 32≈ . units
15. exterior angles must add to 360º.
since there 10, each exterior
angle mustt have a measure
of 360º10
36º. each interioor rr
angle, therefore will have a
measure of 180º −− =36 144º º.
16. arc Bac has a measure of 264º,
so arc Bc has a measure of
360º the me− =264 96º º. aasure
of the inscribed angle is half the
measurre of the arc,
so m Bac∠ = =
+ =
12
96 48
3
5
7
2
3 5
5 5
x º º
17. ++ =
+ = + =
( )( ) +
( )( ) = +
7 2
2 2
3 5
25
7 2
4
3 55
7 22
2 3 52 5
5 7 25 2
6 5 335 210
1 450 000 1 45 106
0076 7 6 10 3
18.
19.
20.
, , .
. .
= ×
= × −
6640 000 000 000 6 4 1011, , , .= ×
Systematic Review 24E1. rH sB≅ (1 and 2 may be reeversed.)
sss postulate
yes; c
2.
3.
4.
5.
rB sH
BH BH
≅
≅
oorresponding angles
are congruent
51º6.
7.
8.
113º
∠∠
+ =
+ =
=
=
tWV
Vt
LL
LL
LL
LL
9.
10. 102 2 202
100 2 4002 300
300
LLL
this
= =100 3 10 3
11. triangle's sides follow the
pattern of 30º
triangles, so a
− −
∠ =
60 90
60
º º
º
12.. this triangle's sides follow the
pattern of 300º
triangles, so b
− −
∠ =
= = (
60 90
30
12
12
10
º º
º
13. a bh )) ( ) =
= + + =
10 3 50 3
86 60
20 10 10 3 30
≈ . square units
14. P ++10 3
47 32≈ . units
15. exterior angles must add to 360º.
since there 10, each exterior
angle mustt have a measure
of 360º10
36º. each interioor rr
angle, therefore will have a
measure of 180º −− =36 144º º.
16. arc Bac has a measure of 264º,
so arc Bc has a measure of
360º the me− =264 96º º. aasure
of the inscribed angle is half the
measurre of the arc,
so m Bac∠ = =
+ =
12
96 48
3
5
7
2
3 5
5 5
x º º
17. ++ =
+ = + =
( )( ) +
( )( ) = +
7 2
2 2
3 5
25
7 2
4
3 55
7 22
2 3 52 5
5 7 25 2
6 5 335 210
1 450 000 1 45 106
0076 7 6 10 3
18.
19.
20.
, , .
. .
= ×
= × −
6640 000 000 000 6 4 1011, , , .= ×
Lesson Practice 25ALesson Practice 25A1.
2.
3.
4.
given
BaD
given
D
∠ ≅ ∠BcD
BB bisects aBc
definition of bisector
aBD
∠
≅
5.
6. cBD
asa
aBD acD
given
aD
7.
8.
9.
10.
11.
yes: cPctrc
∠ ≅ ∠
secbi ts Bac
aD
∠
≅
12.
13.
definition of bisector
aD
114.
15.
16.
17.
aDB aDc
aas
BD
≅
yes: cPctrc
bisects aBc and aDc
18 and 19 may be
∠ ∠
∠ ≅ ∠18. aBD cBD
( reversed)
19.
20.
21.
2
∠ ≅ ∠
≅
≅
aDB cDB
BD BD
BaD BcD
22.
23.
24.
∠ ≅ ∠
≅
BaD BcD
eF eG
e is the midpoint of FG
225.
26.
27.
definition of midpoint
DG HF
DGe H
||
∠ ≅ ∠ FFe
DeG HeF
DeG HeF
GDe FHe
28.
29.
30.
∠ ≅ ∠
≅
∠ ≅ ∠
Lesson Practice 25BLesson Practice 25B1.
2.
3.
4.
given
aB ≅ cD
aB cD
a
||
BBc aDc
BD
given
≅
∠
5.
6.
7.
bisects aBc
definition off bisector
BaD8.
9.
10.
11.
∠ ≅ ∠
≅
≅
BcD
BD BD
aBD cBD
ye
ss cPctrc
given
;
alternate interior angles
12.
13.
144.
15.
16.
17.
given
HG
definition of bisector
HG
≅
∠
FG
MM
yes: cPctrc
given
al
≅ ∠
≅
FGe
HGM FGe18.
19.
20.
21.
tternate interior angles
BaF and eDc are ri22. ∠ ∠ gght angles
BaF23.
24.
25.
26.
∠ ≅ ∠
≅
≅
eDc
ce FB
cDe FaB
aaas
27. yes: cPctrc
Lesson Practice 25B - sYsteMatic reVieW 25c
soLUtions GeoMetrY188
Lesson Practice 25B1.
2.
3.
4.
given
aB ≅ cD
aB cD
a
||
BBc aDc
BD
given
≅
∠
5.
6.
7.
bisects aBc
definition off bisector
BaD8.
9.
10.
11.
∠ ≅ ∠
≅
≅
BcD
BD BD
aBD cBD
ye
ss cPctrc
given
;
alternate interior angles
12.
13.
144.
15.
16.
17.
given
HG
definition of bisector
HG
≅
∠
FG
MM
yes: cPctrc
given
al
≅ ∠
≅
FGe
HGM FGe18.
19.
20.
21.
tternate interior angles
BaF and eDc are ri22. ∠ ∠ gght angles
BaF23.
24.
25.
26.
∠ ≅ ∠
≅
≅
eDc
ce FB
cDe FaB
aaas
27. yes: cPctrc
Systematic Review 25CSystematic Review 25C1. ∠ ≅ ∠QPt QsU
(1 and 2 may be reversed)
Q is the midpoint of Ut
3 a
2.
3. UQ tQ≅
( nnd 5 may be reversed)
definition of midpoint4.
(44 and 6 may be reversed if
3 and 5 were reverseed)
5. ∠ ≅ ∠UQs tQP
6.
7.
8.
9.
vertical angles
yes: cPctrc
QsU QPt
aas
≅
110. 180 123 57º º º− =
11.
12.
180 111 69
180 57 69
180 126 54
º º º
º º º
º º
− =
− +( ) =
− = ºº
13.
14.
15.
Qrs
rs
if the square is viewed as two
ccongruent triangles, one pointing
up, and one ppointing down, we
can find the area of these twwo
triangles, and add them to find
the area of tthe square. since the
diagonals are 12 inches llong, and
bisect each other, we know that
the heeight of each triangle is 6
inches, so the areaa of one
triangle would be:
12
bh i= ( ) ( ) =12
12 6 36 nn
in
2
36
area of both triangles combined:
36 in2 + 22 72 2= in
in problems 16 , your answer
may be slightly
− 20
different from the
one given, depending upon
wwhether the intermediate steps
were rounded or not. answers
that are close may be considered
correct.
Looking at one of the four small
t
16.
rriangles, each of the legs has a
length of 6 inn, so the hypotenuse
of these triangles is:
62 + 662 2
36 36 2
72 2
72 6 2 8 49
4 8 49
=
+ =
=
=
= ( ) =
H
H
H
H or in
P
.
.
≈
333 96
2 3 14 82
200 96
.
. .
in
r
17. area of circle:
π ≈ ( ) = iin
of
i
2
36360
110
200 96
sector the circle
110
= =ºº
. nn in
c
2 20 096 2
2
( ) =
=
.
18. circumference of circle:
πrr in
of
≈ 2 3 14 8 50 24
36360
110
. .
ºº
( ) ( ) =
= =sector the circle
1102 m
50 24 5 024
200
200
. .( ) =
=
= ( )
in
cm
V
19.
3340 560
38 080 000 3
340 3 4
( ) ( ) =
=
, ,
.
cm
or
cm m and 5560 5 6
2 3 4 5 6 38 08 3
2 2
.
. . .
cm m
V m
sa
=
= ( ) ( ) ( ) =
=
( )20.
33 4 2 2 5 6 2 3 4 5 6
13 6 22 4 38 08 74
. . . .
. . .
( ) + ( ) ( ) + ( ) ( ) =
+ + = ..08 2
2 200 340 2 200 560 2 340 560
m
or
sa =
( )( )+ ( )( )+ ( )( ) ==
+ + =136 000 224 000 380 800
740 800 2
, , ,
, cm
in problems 16 , your answer
may be slightly
− 20
different from the
one given, depending upon
wwhether the intermediate steps
were rounded or not. answers
that are close may be considered
correct.
Looking at one of the four small
t
16.
rriangles, each of the legs has a
length of 6 inn, so the hypotenuse
of these triangles is:
62 + 662 2
36 36 2
72 2
72 6 2 8 49
4 8 49
=
+ =
=
=
= ( ) =
H
H
H
H or in
P
.
.
≈
333 96
2 3 14 82
200 96
.
. .
in
r
17. area of circle:
π ≈ ( ) = iin
of
i
2
36360
110
200 96
sector the circle
110
= =ºº
. nn in
c
2 20 096 2
2
( ) =
=
.
18. circumference of circle:
πrr in
of
≈ 2 3 14 8 50 24
36360
110
. .
ºº
( ) ( ) =
= =sector the circle
1102 m
50 24 5 024
200
200
. .( ) =
=
= ( )
in
cm
V
19.
3340 560
38 080 000 3
340 3 4
( ) ( ) =
=
, ,
.
cm
or
cm m and 5560 5 6
2 3 4 5 6 38 08 3
2 2
.
. . .
cm m
V m
sa
=
= ( ) ( ) ( ) =
=
( )20.
33 4 2 2 5 6 2 3 4 5 6
13 6 22 4 38 08 74
. . . .
. . .
( ) + ( ) ( ) + ( ) ( ) =
+ + = ..08 2
2 200 340 2 200 560 2 340 560
m
or
sa =
( )( )+ ( )( )+ ( )( ) ==
+ + =136 000 224 000 380 800
740 800 2
, , ,
, cm
sYsteMatic reVieW 25c - sYsteMatic reVieW 25D
soLUtionsGeoMetrY 189
in problems 16 , your answer
may be slightly
− 20
different from the
one given, depending upon
wwhether the intermediate steps
were rounded or not. answers
that are close may be considered
correct.
Looking at one of the four small
t
16.
rriangles, each of the legs has a
length of 6 inn, so the hypotenuse
of these triangles is:
62 + 662 2
36 36 2
72 2
72 6 2 8 49
4 8 49
=
+ =
=
=
= ( ) =
H
H
H
H or in
P
.
.
≈
333 96
2 3 14 82
200 96
.
. .
in
r
17. area of circle:
π ≈ ( ) = iin
of
i
2
36360
110
200 96
sector the circle
110
= =ºº
. nn in
c
2 20 096 2
2
( ) =
=
.
18. circumference of circle:
πrr in
of
≈ 2 3 14 8 50 24
36360
110
. .
ºº
( ) ( ) =
= =sector the circle
1102 m
50 24 5 024
200
200
. .( ) =
=
= ( )
in
cm
V
19.
3340 560
38 080 000 3
340 3 4
( ) ( ) =
=
, ,
.
cm
or
cm m and 5560 5 6
2 3 4 5 6 38 08 3
2 2
.
. . .
cm m
V m
sa
=
= ( ) ( ) ( ) =
=
( )20.
33 4 2 2 5 6 2 3 4 5 6
13 6 22 4 38 08 74
. . . .
. . .
( ) + ( ) ( ) + ( ) ( ) =
+ + = ..08 2
2 200 340 2 200 560 2 340 560
m
or
sa =
( )( )+ ( )( )+ ( )( ) ==
+ + =136 000 224 000 380 800
740 800 2
, , ,
, cm
Systematic Review 25DSystematic Review 25DKeep in mind the fact that the order
of some steps in these proofs may
be iinterchangeable.
bisects JKX
bisects
1.
2.
KZ
KZ
∠
∠∠
∠ ≅ ∠
∠ ≅
XZJ
definition of bisector
3.
4.
5.
JKZ XKZ
XZK ∠∠JZK
6.
7.
8.
definition of bisector
reflexive prop
KZ KZ≅
eerty
JKZ9.
10.
≅ XKZ
asa
11.
12.
13.
KJ KX≅
− =
− +( ) =
180 83 97
180 28 97
180
º º º
º º º
º −− =
+ =
+ =
=
= = =
125 55
2 62 152
2 36 225
2 189
189 9 21
º º
14. s
s
s
s 33 21 13 75
6
≈ .
15. Base of top triangle
Height of
= cm
top triangle
area of top triangle =
= ( ) =12
8
4 cm
12
top and bottom triangles
are co
6 4
12 2
( ) ( ) =
cm
nngruent, so:
a = + =12 12 24 2cm
11.
12.
13.
KJ KX≅
− =
− +( ) =
180 83 97
180 28 97
180
º º º
º º º
º −− =
+ =
+ =
=
= = =
125 55
2 62 152
2 36 225
2 189
189 9 21
º º
14. s
s
s
s 33 21 13 75
6
≈ .
15. Base of top triangle
Height of
= cm
top triangle
area of top triangle =
= ( ) =12
8
4 cm
12
top and bottom triangles
are co
6 4
12 2
( ) ( ) =
cm
nngruent, so:
a = + =12 12 24 2cm
16. the four small triangles each have
sides meassuring 4 cm and 3 cm, so
the hypotenuse of eachh is:
H
H
H
H
Perimeter
2 42 32
2 16 9
2 25
5
4 5 20
= +
= +
=
=
= ( ) = ccm
17. Volume of cylinder:
V Bh r h= =
( ) ( )
π ≈
≈
2
3 14 42
3 3 1. . 665 79 3
10360
136
165
.
ºº
cm
ofsection cylinder
136
= =
.. .79 3 4 61 3cm cm( ) ≈
18. circumference of circle:
c r= ( ) ( ) =2 2 3 14 4 25π ≈ . .112
10360
136
136
25 12 70
ºº º
. .
cm
arc of circle= =
( ) ≈ ccm
19. convert all measurements
to yards:
61
1ft × yyd
yd
3
63
2
3 31
1
3
3 33
31
3
ft
yd yd
ft
ft
yd
yd y
= =
× = =
= dd
volume rea of base times
height. triangular
= a
eend is base, so:
V Bh= = ( )( ) ( )
=
12
2 3 5
5 3 8 66 3≈ . yd
20.. sa = area of three rectangular sides,
plus area of two triangular ends.
note that the ends aree equilateral
triangles. this can be verified bby
using the pythagorean theorem
and the informmation given.
sa =
( )( ) + (3 5 2 2 12
2yd yd ( ) yd))( ) =
+ =
+
3
30 2 2 3 2
30 2 3 2 33 46 2
yd
yd yd
yd . yd≈
sYsteMatic reVieW 25D - sYsteMatic reVieW 25e
soLUtions GeoMetrY190
18. circumference of circle:
c r= ( ) ( ) =2 2 3 14 4 25π ≈ . .112
10360
136
136
25 12 70
ºº º
. .
cm
arc of circle= =
( ) ≈ ccm
19. convert all measurements
to yards:
61
1ft × yyd
yd
3
63
2
3 31
1
3
3 33
31
3
ft
yd yd
ft
ft
yd
yd y
= =
× = =
= dd
volume rea of base times
height. triangular
= a
eend is base, so:
V Bh= = ( )( ) ( )
=
12
2 3 5
5 3 8 66 3≈ . yd
20.. sa = area of three rectangular sides,
plus area of two triangular ends.
note that the ends aree equilateral
triangles. this can be verified bby
using the pythagorean theorem
and the informmation given.
sa =
( )( ) + (3 5 2 2 12
2yd yd ( ) yd))( ) =
+ =
+
3
30 2 2 3 2
30 2 3 2 33 46 2
yd
yd yd
yd . yd≈
Systematic Review 25ELesson Practice 25E1.
2.
∠ ≅ ∠LPG rGP
alternate interiior angles
rPG
alternate interior angl
3.
4.
∠ ≅ ∠LGP
ees
5. PG PG≅
6.
7.
8.
9.
reflexive property
PLG
yes: cPc
≅ GrP
asa
ttrc
exterior angles of a polygon
add up to 3
10.
660º:
360 115 119
360 234 126
º º º
º º º
− +( ) =
− =
11.
12.
180 115 65
180 126 54
º º º
º º º
− =
− =
(1226 from answer #10)
top tri
º
º º º13.
14.
180 119 61− =
aangle:
Bottom triangle:
a bh in
a
= = ( ) ( ) =12
12
10 6 30 2
== = ( ) ( ) =
+
12
12
10 10 50 2
50
bh in
in
total area:
30 in2 22 80 2
62 52
=
= +
in
15. one of two small triangles:
h2
hh
h
h in
2 36 25
2 61
61 7 81
= +
=
= ≈ .
one of two large trianngles:
h2 = +
= +
=
=
=
102 52
2 100 25
2 125
125 11 18
h
h
h in
P
≈ .
22 7 81 2 11 18
15 62 22 36 37 98
. .
. . .
( ) + ( ) =
+ = in
16. circle
a r mm
:
. .
º
= ( ) =
=
π ≈2 3 14 72
153 86 2
2403
sector660
23
23
153 86 2 102 57 2º
. .
=
( )of circle
mm mm
ci
≈
17. rrcle
c r mm
:
. .
ºº
= ( ) ( ) =
= =
2 2 3 14 7 43 96
240360
π ≈
sector 223
23
43 96 29 31. .
of circle
mm mm( ) ≈
18. 5 000 8 000 000
5 103 8 106
5 8 103 106
, , ,× =
×( ) ×( ) =
×( ) ×( )) =
× = ×40 109 4 1010
sYsteMatic reVieW 25e - Lesson Practice 26B
soLUtionsGeoMetrY 191
19. 18 000 007
1 8 7 0 104 10 3
12 6 101
, .
. .
.
× =
×( ) × −( ) =
( ) ( ) = 112 6 101
1 26 102
1 102
.
.
× =
×
×or:
if the student took significant
digits into account.
20. 1 400 000 2, , ÷ 990
1 4 106 2 9 102
1 4 2 9 106 102
48
=
×( ) ×( ) =
( ) ( ) =
×
. .
. .
.
÷
÷ ÷
1104 4 8 103= ×.
Lesson Practice 26ALesson Practice 26A1.
2.
ac BD⊥
definition of perpenndicular
definition of perpendicular
aBD is
3.
4. isosceles
aB5.
6.
7.
8.
9.
≅
≅
≅
aD
ac ac
aBc aDc
cPctrc
cc is the midpoint of
definition of midpoi
BD
10. nnt
HJM is equilateral
angles in an equila
11.
12.
tteral
triangle are congruent
definit
13.
14.
HK JM⊥
iion of perpendicular
definition of perpendic15. uular
HJK
aBcD is a rhombus
16.
17.
18.
1
HK HK
HMK
≅
≅
99.
20.
21.
22.
aB
given
aoB is a right angl
≅
⊥
∠
cD
ac DB
ee
coD is a right angle
is the midpoint
23.
24.
∠
o oof BD
definition of midpoint
aBo
25.
26.
27.
oD oB≅
≅ cDo
HL28.
Lesson Practice 26A1.
2.
ac BD⊥
definition of perpenndicular
definition of perpendicular
aBD is
3.
4. isosceles
aB5.
6.
7.
8.
9.
≅
≅
≅
aD
ac ac
aBc aDc
cPctrc
cc is the midpoint of
definition of midpoi
BD
10. nnt
HJM is equilateral
angles in an equila
11.
12.
tteral
triangle are congruent
definit
13.
14.
HK JM⊥
iion of perpendicular
definition of perpendic15. uular
HJK
aBcD is a rhombus
16.
17.
18.
1
HK HK
HMK
≅
≅
99.
20.
21.
22.
aB
given
aoB is a right angl
≅
⊥
∠
cD
ac DB
ee
coD is a right angle
is the midpoint
23.
24.
∠
o oof BD
definition of midpoint
aBo
25.
26.
27.
oD oB≅
≅ cDo
HL28.
Lesson Practice 26BLesson Practice 26B1.
2.
3.
He Fc
BG Fc
BGc
⊥
⊥
∠ is a rigght angle
eHF is a right angle
He
given
4.
5.
6.
∠
≅ GB
77.
8.
9.
10.
11.
Fe
given
ac is tangen
≅
≅
cB
FHe cGB
HL
tt to circle at B
DBa is a right angle
pro
12.
13.
∠
pperty of tangent
see lesson 12
is a r
( )∠14. DBc iight angle
property of tangent
B is the mi
15.
16. ddpoint of ac
definition of midpoint
17.
18.
1
aB cB≅
99.
20.
21.
22.
DB
reflexive property
aBD
LL
≅
≅
DB
cBD
223.
24.
aG is tangent to circle at G
De is tangentt to circle at e
aG
property of tangent
25.
26.
⊥ GF
227.
28.
29.
De
property of tangent
aGF is a rig
⊥
∠
eF
hht angle
DeF is a right angle
GaF
30.
31.
3
∠
∠ ≅ ∠eDF
22.
33.
34.
given
GF
f two line segments have e
≅ eF
i qqual
lengths, they are congruent.
radii of a ciircle have equal
lengths.
aGF35.
36.
37.
≅ DeF
La
aaF DF
cPctrc
≅
38.
Lesson Practice 26B - sYsteMatic reVieW 26D
soLUtions GeoMetrY192
Lesson Practice 26B1.
2.
3.
He Fc
BG Fc
BGc
⊥
⊥
∠ is a rigght angle
eHF is a right angle
He
given
4.
5.
6.
∠
≅ GB
77.
8.
9.
10.
11.
Fe
given
ac is tangen
≅
≅
cB
FHe cGB
HL
tt to circle at B
DBa is a right angle
pro
12.
13.
∠
pperty of tangent
see lesson 12
is a r
( )∠14. DBc iight angle
property of tangent
B is the mi
15.
16. ddpoint of ac
definition of midpoint
17.
18.
1
aB cB≅
99.
20.
21.
22.
DB
reflexive property
aBD
LL
≅
≅
DB
cBD
223.
24.
aG is tangent to circle at G
De is tangentt to circle at e
aG
property of tangent
25.
26.
⊥ GF
227.
28.
29.
De
property of tangent
aGF is a rig
⊥
∠
eF
hht angle
DeF is a right angle
GaF
30.
31.
3
∠
∠ ≅ ∠eDF
22.
33.
34.
given
GF
f two line segments have e
≅ eF
i qqual
lengths, they are congruent.
radii of a ciircle have equal
lengths.
aGF35.
36.
37.
≅ DeF
La
aaF DF
cPctrc
≅
38.
Systematic Review 26CSystematic Review 26C1.
2.
rL Gn or rn GL≅ ≅
opposite sides of a rectangle
are congruent (aPt)
3. m n∠ rrL m LGn
L
≅ ∠ =
≅
º90
4.
5.
definition of a rectangle
Ln nn
GLn
HL
6.
7.
8.
9.
reflexive property
rnL
L2
≅
+ ( )6 222
122
2 6 6 2 2 144
2 36 2 144
2 72 14
=
+ ( ) ( ) ( )( ) =
+ ( ) ( ) =
+ =
L
L
L 442 72
72
36 2
6 2
8 49
62 92 2
3
L
L
L
L
or units
H
=
=
=
=
+ =
.≈
10.
66 81 2
117 2
117 13
10 82
18
+ =
=
=
H
H
H or
or units.
3
≈
11. 00
2
2
2
5
5
2
10
10
10
10
10
10
10 10
100
10 10
º
12.
13.
14. x = =
× = =
11010
2 2 3 14 2 3 14 44
2 3
=
= ( ) ( )
=
15.
16.
c r cm
a r
π ≈ ≈
π ≈
. . .
.114 2 32
16 61 2( ) ( ). .≈ cm
e17. xterior angles add up tto 360º,
so the measure of each exterior
angle iss 360º15
interior angles
p
=
= − =
24
180 24 156
º.
º º º
18. ooint
123º
lternate exterior angles are co
19.
20. a nngruent.
Systematic Review 26C1.
2.
rL Gn or rn GL≅ ≅
opposite sides of a rectangle
are congruent (aPt)
3. m n∠ rrL m LGn
L
≅ ∠ =
≅
º90
4.
5.
definition of a rectangle
Ln nn
GLn
HL
6.
7.
8.
9.
reflexive property
rnL
L2
≅
+ ( )6 222
122
2 6 6 2 2 144
2 36 2 144
2 72 14
=
+ ( ) ( ) ( )( ) =
+ ( ) ( ) =
+ =
L
L
L 442 72
72
36 2
6 2
8 49
62 92 2
3
L
L
L
L
or units
H
=
=
=
=
+ =
.≈
10.
66 81 2
117 2
117 13
10 82
18
+ =
=
=
H
H
H or
or units.
3
≈
11. 00
2
2
2
5
5
2
10
10
10
10
10
10
10 10
100
10 10
º
12.
13.
14. x = =
× = =
11010
2 2 3 14 2 3 14 44
2 3
=
= ( ) ( )
=
15.
16.
c r cm
a r
π ≈ ≈
π ≈
. . .
.114 2 32
16 61 2( ) ( ). .≈ cm
e17. xterior angles add up tto 360º,
so the measure of each exterior
angle iss 360º15
interior angles
p
=
= − =
24
180 24 156
º.
º º º
18. ooint
123º
lternate exterior angles are co
19.
20. a nngruent.
Systematic Review 26DSystematic Review 26D1.
2.
Pt Ps≅
definition of midppoint
PQ
reflexive property
PtQ
3.
4.
5.
6.
≅
≅
PQ
PsQ
LLL
sQ tQ
cPctrc
7.
8.
9.
≅
the perpendicular bisector wwill
divide the triangle into two smaller
trianngles, each of which will have
a base of 4 inchhes, because of the
definition of bisector. Usiing one
these small triangles, we have
a le
of
gg of 4 and a hypotenuse of 8.
Using L for the uunknown leg
L
L
L
L
L
L
:
2 42 82
2 16 642 48
48
16 3
4 3
+ =
+ =
=
=
=
= in
a bh
in
10.
11.
= = ( ) ( ) =
( )( ) =
12
12
8 4 3
4 4 3 16 3 2
each oof the smaller triangles are
45º − −45 90º º trianglles, so they
both have a pair of legs with
equall measures. therefore, the
base of the larger ttriangle has a
measure of 7 + 7
the
= 14 .inches
hhypotenuse of each of the
smaller triangles is 2 times the
leg, or 7 2.
P s s s= + + = + + =
+
14 7 2 7 2
14 14 22 33 80
12
12
14 7 49 2
8
5
3 2
≈ . in
a bh in12.
13.
14.
= = ( )( ) =
×× = =
( ) =
= ( ) ( )
3
4 2
15
12 4
1512 2
1524
43
3 43
3 14 43
15. V rπ ≈ . ≈≈
π ≈
267 95 3
4 2 4 3 14 42
200 96 2
.
.
.
in
a r
in
16.
17
= ( ) ( )
=
.. exterior angles add up to 360º,
so the measuree of each exterior
angle º.
interior a
= =36018
20º
nngles
line or line segment or
= − =180 20 160º º º
18. rray
m a
hey are supplementa
19.
20.
∠ = − =180 72 108º º º
t rry angles.
sYsteMatic reVieW 26D - Lesson Practice 27a
soLUtionsGeoMetrY 193
Systematic Review 26D1.
2.
Pt Ps≅
definition of midppoint
PQ
reflexive property
PtQ
3.
4.
5.
6.
≅
≅
PQ
PsQ
LLL
sQ tQ
cPctrc
7.
8.
9.
≅
the perpendicular bisector wwill
divide the triangle into two smaller
trianngles, each of which will have
a base of 4 inchhes, because of the
definition of bisector. Usiing one
these small triangles, we have
a le
of
gg of 4 and a hypotenuse of 8.
Using L for the uunknown leg
L
L
L
L
L
L
:
2 42 82
2 16 642 48
48
16 3
4 3
+ =
+ =
=
=
=
= in
a bh
in
10.
11.
= = ( ) ( ) =
( )( ) =
12
12
8 4 3
4 4 3 16 3 2
each oof the smaller triangles are
45º − −45 90º º trianglles, so they
both have a pair of legs with
equall measures. therefore, the
base of the larger ttriangle has a
measure of 7 + 7
the
= 14 .inches
hhypotenuse of each of the
smaller triangles is 2 times the
leg, or 7 2.
P s s s= + + = + + =
+
14 7 2 7 2
14 14 22 33 80
12
12
14 7 49 2
8
5
3 2
≈ . in
a bh in12.
13.
14.
= = ( )( ) =
×× = =
( ) =
= ( ) ( )
3
4 2
15
12 4
1512 2
1524
43
3 43
3 14 43
15. V rπ ≈ . ≈≈
π ≈
267 95 3
4 2 4 3 14 42
200 96 2
.
.
.
in
a r
in
16.
17
= ( ) ( )
=
.. exterior angles add up to 360º,
so the measuree of each exterior
angle º.
interior a
= =36018
20º
nngles
line or line segment or
= − =180 20 160º º º
18. rray
m a
hey are supplementa
19.
20.
∠ = − =180 72 108º º º
t rry angles.
Systematic Review 26ESystematic Review 26E1.
2.
∠ ≅ ∠rQs tQs
definition of bisector
Qs
reflexive property
rsQ
3.
4.
5.
≅
≅
Qs
t ssQ
La
rQ tQ
cPctrc
6.
7.
8.
9.
≅
convert 30 cm to m:
301000
=
= = ( ) ( )
= +
.
. .
.
3
2 3 14 32
4
1 13 3
2 2 2
V Bh r h
m
sa r
π ≈ ≈
π10. ππ ≈
≈
rh
2 3 14 32
2 3 14 3 4
57 7 54 8 11
. . . .
. . .
( ) ( ) + ( ) ( ) ( )
+ = mm2
Your answer may differ slightly
from this, ddepending upon how
and when you rounded.
sid11. ee-side-side
side-angle-side
angle-side-ang
12.
13. lle
angle-angle-side
hypotenuse-leg
leg-l
14.
15.
16. eeg
hypotenuse-angle
leg-angle
17.
18.
19. a r= ≅π 2 3 14. 222
3 14 2 2
3 14 4 2 12 56 2 2
X X X
X X units
( ) = ( )( ) =
( ) =
.
. .
20. 222
32 2
4 2 9 2 2
13 2 2
13 2
2 13 1
X X H
X X H
X H
X H
H X X
( ) + ( ) =
+ =
=
=
= = 33 units
Systematic Review 26E1.
2.
∠ ≅ ∠rQs tQs
definition of bisector
Qs
reflexive property
rsQ
3.
4.
5.
≅
≅
Qs
t ssQ
La
rQ tQ
cPctrc
6.
7.
8.
9.
≅
convert 30 cm to m:
301000
=
= = ( ) ( )
= +
.
. .
.
3
2 3 14 32
4
1 13 3
2 2 2
V Bh r h
m
sa r
π ≈ ≈
π10. ππ ≈
≈
rh
2 3 14 32
2 3 14 3 4
57 7 54 8 11
. . . .
. . .
( ) ( ) + ( ) ( ) ( )
+ = mm2
Your answer may differ slightly
from this, ddepending upon how
and when you rounded.
sid11. ee-side-side
side-angle-side
angle-side-ang
12.
13. lle
angle-angle-side
hypotenuse-leg
leg-l
14.
15.
16. eeg
hypotenuse-angle
leg-angle
17.
18.
19. a r= ≅π 2 3 14. 222
3 14 2 2
3 14 4 2 12 56 2 2
X X X
X X units
( ) = ( )( ) =
( ) =
.
. .
20. 222
32 2
4 2 9 2 2
13 2 2
13 2
2 13 1
X X H
X X H
X H
X H
H X X
( ) + ( ) =
+ =
=
=
= = 33 units
Lesson Practice 27ALesson Practice 27A1.
2.
3.
4.
aB BD
ec BD
aBD ecD
⊥
⊥
∠ ≅ ∠
deefinition of perpendicular
eDc
reflexi
5.
6.
∠ ≅ ∠aDB
vve property
aBe
7.
8.
9.
10.
aBD ecD
aa
aB cD
c
~
||
∠ ≅ ∠ DDe
ceD
11.
12.
13.
alternate interior angles
aeB
v
∠ ≅ ∠
eertical angles
For problems u
14.
15.
aBe cDe
aa
~
ssing proportions,
there is often more than one
possible way to set up the
proportion. the stuudent can use
any method that results in a
correect answer.
16.
17.
510
8
5 80
16
1015 2424015
=
=
=
=
=
XX
X
X
XX
X
XX
X
X
X
X
=
=
=
=
=
=
=
16
1525
6
15 150
10
810 1310410
1
18.
19.
00 4
5 610
50 6
506
8 13
.
20.X
X
X
=
=
= =
Lesson Practice 27a - sYsteMatic reVieW 27c
soLUtions GeoMetrY194
Lesson Practice 27A1.
2.
3.
4.
aB BD
ec BD
aBD ecD
⊥
⊥
∠ ≅ ∠
deefinition of perpendicular
eDc
reflexi
5.
6.
∠ ≅ ∠aDB
vve property
aBe
7.
8.
9.
10.
aBD ecD
aa
aB cD
c
~
||
∠ ≅ ∠ DDe
ceD
11.
12.
13.
alternate interior angles
aeB
v
∠ ≅ ∠
eertical angles
For problems u
14.
15.
aBe cDe
aa
~
ssing proportions,
there is often more than one
possible way to set up the
proportion. the stuudent can use
any method that results in a
correect answer.
16.
17.
510
8
5 80
16
1015 2424015
=
=
=
=
=
XX
X
X
XX
X
XX
X
X
X
X
=
=
=
=
=
=
=
16
1525
6
15 150
10
810 1310410
1
18.
19.
00 4
5 610
50 6
506
8 13
.
20.X
X
X
=
=
= =
Lesson Practice 27BLesson Practice 27B1.
2.
3.
∠ ≅ ∠
∠ ≅ ∠
DaB eBc
acD Bce
refllexive property
gi
4.
5.
6.
7.
Dac eBc
aa
aBc aDB
~
∠ ≅ ∠
vven
reflexive property
8.
9.
10.
∠ ≅ ∠DaB Bac
aBc aDB ~
111.
12.
13.
aa
X
X
X or
X
X
1058
8 50
508
6 14
6 25
496
6
=
=
= =
=
.
==
=
=
=
= =
36
6
153325
25 495
49525
19 45
19 8
X
X
X
X or
14.
15
.
..
16.
X
X
X or
X
X
136
1515 78
7815
5 15
5 2
61510
10 9
=
=
= =
=
=
.
00
9
55
1610
10 5 80
10 50 80
10 30
3
X
X
X
X
X
X
=
+( )=
+( ) =
+ =
=
=
17.
Lesson Practice 27B1.
2.
3.
∠ ≅ ∠
∠ ≅ ∠
DaB eBc
acD Bce
refllexive property
gi
4.
5.
6.
7.
Dac eBc
aa
aBc aDB
~
∠ ≅ ∠
vven
reflexive property
8.
9.
10.
∠ ≅ ∠DaB Bac
aBc aDB ~
111.
12.
13.
aa
X
X
X or
X
X
1058
8 50
508
6 14
6 25
496
6
=
=
= =
=
.
==
=
=
=
= =
36
6
153325
25 495
49525
19 45
19 8
X
X
X
X or
14.
15
.
..
16.
X
X
X or
X
X
136
1515 78
7815
5 15
5 2
61510
10 9
=
=
= =
=
=
.
00
9
55
1610
10 5 80
10 50 80
10 30
3
X
X
X
X
X
X
=
+( )=
+( ) =
+ =
=
=
17.
Systematic Review 27CSystematic Review 27C1.
2.
∠ ≅ ∠FGH FJK
definition of perpendicular
GFH
reflexive property
3.
4.
5
∠ ≅ ∠JFK
..
6.
7.
FGH FJK
aa
rct and etc
~
∠ ∠
are right angles
88.
9.
10.
definition of rectangle
rc
definition
≅ et
oof rectangle
ct
reflexive property
11.
12.
13.
≅ ct
e cct rtc
X
X
X
≅
=
=
14.
15.
16.
LL
yes: cPctrc
101525
25 150
==
=
6
10
17. exterior angles add up to 360º
360º36º
ssides: decagon
a decagon can be divided into18. 10
congruent triangles. the area of
one of theese triangles would be:
a bh c= = ( ) ( ) =12
12
5 3 2 7 5 2. mm
triangles cm
2
7 5 2 10 75 2 2
106 07
18
.
.
( ) =
≈ cm2
19. 00 105 75
180 4
º º º:
º
− =
∠ = −
supplementary angles
20. m α 33 75
180 118 62
º º
º º º
+( ) =
− =
sYsteMatic reVieW 27c - sYsteMatic reVieW 27e
soLUtionsGeoMetrY 195
Systematic Review 27C1.
2.
∠ ≅ ∠FGH FJK
definition of perpendicular
GFH
reflexive property
3.
4.
5
∠ ≅ ∠JFK
..
6.
7.
FGH FJK
aa
rct and etc
~
∠ ∠
are right angles
88.
9.
10.
definition of rectangle
rc
definition
≅ et
oof rectangle
ct
reflexive property
11.
12.
13.
≅ ct
e cct rtc
X
X
X
≅
=
=
14.
15.
16.
LL
yes: cPctrc
101525
25 150
==
=
6
10
17. exterior angles add up to 360º
360º36º
ssides: decagon
a decagon can be divided into18. 10
congruent triangles. the area of
one of theese triangles would be:
a bh c= = ( ) ( ) =12
12
5 3 2 7 5 2. mm
triangles cm
2
7 5 2 10 75 2 2
106 07
18
.
.
( ) =
≈ cm2
19. 00 105 75
180 4
º º º:
º
− =
∠ = −
supplementary angles
20. m α 33 75
180 118 62
º º
º º º
+( ) =
− =
Systematic Review 27DSystematic Review 27D1.
2.
∠ ≅ ∠aeB DcB
alternate inteerior angles
aBe
vertical angles
3.
4.
5.
∠ ≅ ∠DBc
aeB ~~DcB
aa
aB cB
BD
6.
7.
8.
9.
10.
≅
≅
given
BD
reflexive propperty
cPctrc
11.
12.
13.
14.
15.
aBD cBD
HL
aD cD
X
≅
≅
=5
33010
10 150
15
X
X
=
=
16. exterior angles add up to 3600º:
: octagon
area of one tria
36045
8ºº
= sides
17. nngle:
area of octagon:
60
a bh m= = ( ) ( ) =12
12
10 12 60 2
m2 8 480 2
8 10 80
180º
triangles m
P m
( ) =
= ( ) =
−
18.
19. 999 81
180 33 8
º º
º º
=
∠ = − +
:
supplementary angles
20. m a 11
180 114 66
º
º º º
( ) =
− =
Systematic Review 27D1.
2.
∠ ≅ ∠aeB DcB
alternate inteerior angles
aBe
vertical angles
3.
4.
5.
∠ ≅ ∠DBc
aeB ~~DcB
aa
aB cB
BD
6.
7.
8.
9.
10.
≅
≅
given
BD
reflexive propperty
cPctrc
11.
12.
13.
14.
15.
aBD cBD
HL
aD cD
X
≅
≅
=5
33010
10 150
15
X
X
=
=
16. exterior angles add up to 3600º:
: octagon
area of one tria
36045
8ºº
= sides
17. nngle:
area of octagon:
60
a bh m= = ( ) ( ) =12
12
10 12 60 2
m2 8 480 2
8 10 80
180º
triangles m
P m
( ) =
= ( ) =
−
18.
19. 999 81
180 33 8
º º
º º
=
∠ = − +
:
supplementary angles
20. m a 11
180 114 66
º
º º º
( ) =
− =
Systematic Review 27ESystematic Review 27E1.
2.
3.
4
∠ ≅ ∠
∠ ≅ ∠
1 2
given
MLn QPr
.. two angles with the
same measure are congruentt
aa
definition of isoscel
5.
6.
7.
8.
LMn PQr
cF cD
~
≅
ees triangle
FcM
definition of bisecto
9.
10.
∠ ≅ ∠DcM
rr
cM
reflexive property
cDM
11.
12.
13.
14.
≅
≅
cM
cFM
ssas
X
X
X
X
15. +( )=
+( ) =
+ =
=
1212
2015
15 12 240
15 180 240
15 660
4
2 3 14 62
113 04
X
a r
=
= ( ) =
16. area of circle:
π ≈ . . ftt
.
2
1112
113 04
sector is 330º360º
of circle
1112
=
fft . ft2 103 62 2
2
( ) =
= =
17. Volume of cylinder:
V Bh rπ hh ≈
3 14 62
4 5 508 68 3
1
. . . ft( ) ( ) =
=sector is 330º360º
1112
508 68 3 466 29 3
2
. ft . ft
of cylinder
1112
( ) =
18. XX X
X
X
º º º
º º
º
+ =
( )=
=
3 180
5 180
36
supplementary angles
119. 4 90
5 90
1
X X
X
X
º º º
º º
+ =
( )=
=
complementary angles
88
2 6 4
2 6 4 12
12 12
1
º
20. P s s s X X X
X X X
X
X
= + + = + +
+ + =
=
=
sYsteMatic reVieW 27e - Lesson Practice 28a
soLUtions GeoMetrY196
Systematic Review 27E1.
2.
3.
4
∠ ≅ ∠
∠ ≅ ∠
1 2
given
MLn QPr
.. two angles with the
same measure are congruentt
aa
definition of isoscel
5.
6.
7.
8.
LMn PQr
cF cD
~
≅
ees triangle
FcM
definition of bisecto
9.
10.
∠ ≅ ∠DcM
rr
cM
reflexive property
cDM
11.
12.
13.
14.
≅
≅
cM
cFM
ssas
X
X
X
X
15. +( )=
+( ) =
+ =
=
1212
2015
15 12 240
15 180 240
15 660
4
2 3 14 62
113 04
X
a r
=
= ( ) =
16. area of circle:
π ≈ . . ftt
.
2
1112
113 04
sector is 330º360º
of circle
1112
=
fft . ft2 103 62 2
2
( ) =
= =
17. Volume of cylinder:
V Bh rπ hh ≈
3 14 62
4 5 508 68 3
1
. . . ft( ) ( ) =
=sector is 330º360º
1112
508 68 3 466 29 3
2
. ft . ft
of cylinder
1112
( ) =
18. XX X
X
X
º º º
º º
º
+ =
( )=
=
3 180
5 180
36
supplementary angles
119. 4 90
5 90
1
X X
X
X
º º º
º º
+ =
( )=
=
complementary angles
88
2 6 4
2 6 4 12
12 12
1
º
20. P s s s X X X
X X X
X
X
= + + = + +
+ + =
=
=
Lesson Practice 28ALesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28a - Lesson Practice 28B
soLUtionsGeoMetrY 197
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28A Here are some ideas to helpp with
transformational geometry. For
reflectioon problems, try placing the
edge of a hand mirrror on the axis
of reflection, facing the figurre to be
reflected. the reflection in the mirroor
will show the image that would be the
result of geometric reflection.
For rotation problemms, lay a piece of
tracing paper over the graphh, trace the
original figure, and plot the poinnt of
rotation. Without moving the tracing
papeer, insert a pin through the point
of rotation and into the graph on the
student page. now, rrotate the tracing
paper counterclockwise the iindicated
number of degrees, and the new
positiion of the image on the tracing
paper will indiicate the result of
geometric rotation.
1.
2.
3.
4..
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28BLesson Practice 28B
1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lession Practice 28B - sYsteMatic reVieW 28c
soLUtions GeoMetrY198
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
XLesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lesson Practice 28B1.
2.
3.
4.
5.
6.
7.
8.
9.
Y
X
Lession Practice 28B
Systematic Review 28CLesson Practice 28C1- 4.
5 - 8.
Y
X
4
2
sYsteMatic reVieW 28c - sYsteMatic reVieW 28D
soLUtionsGeoMetrY 199
Lesson Practice 28C1- 4.
5 - 8.
Y
X
6 8
9.
10.
11.
D a cD
a c
DD
D
: aB
= +
+ = + =
=
=
=
5 10 15
53
15
5 45
9
ft
fft
ft
12.
13.
14.
610 30180 10
18
6 2
=
=
=
X
X
X
reflex angle
000 6 100 2
6 100 2 6 10 2 60 2
= ( )( ) =
= ( ) =
∠ = ∠15.
16.
m Y m P
givenn
angles with the same
measure are
17.
18.
m X m n∠ = ∠
congruent
19.
20.
WXY MnP
aa
~
Systematic Review 28C
Systematic Review 28DLesson Practice 28C1- 4.
5 - 8.
2
4
y = –1
Systematic Review 28D1.
2.
3.
4.
5.
6. B
B
15 646
6 4
+( ) =
= 115 6
6 4 21
6 84
14
15 104
10
10 4 1
+( )= ( )=
=
+( ) =
=
B
B
B
B
B
ft
7.
55 10
10 4 25
10 100
10
102 252 2
100 6
+( )= ( )=
=
+ =
+
B
B
B
H
ft
8.
225 2
725 2
725
25
26 93
1
=
=
=
=
∠ =
H
H
H
H
H
m
29 = 5 29
≈ . ft
9. α 880 39 44
180 83 97
180 97 83
º º º
º º º
º º º
− +( ) =
− =
∠ = − =10. m β
ssupplementary angles
isosc
( )11. right triangle, eeles
major arc and minor arc add up
to 360º:
12.
3360 79 281
120
º º º
º
− =
− =
13. exterior angles are
180º 660
36060
6
º
ºº
exterior angles add up to 360º:
= sidd
bh
es: hexagon
area of one triangle14. =
=12
12
7 5.(( ) ( ) =
( ) =
5 18 75 2
112
.
.
in
six triangles:
6 18.75 in2 55 2
6 7 5 45.
in
P in in
rHsB
15.
16.
= ( ) =
is a rhombus
117.
18.
Bs definition of a rhombus
def
≅
≅
rH
Br Hs
:
: iinition of a rhombus
reflexive proper19. BH BH≅ : tty
HsB sss20. ≅ :BrH
4 ft
15 ft
Systematic Review 28D1.
2.
3.
4.
5.
6. B
B
15 646
6 4
+( ) =
= 115 6
6 4 21
6 84
14
15 104
10
10 4 1
+( )= ( )=
=
+( ) =
=
B
B
B
B
B
ft
7.
55 10
10 4 25
10 100
10
102 252 2
100 6
+( )= ( )=
=
+ =
+
B
B
B
H
ft
8.
225 2
725 2
725
25
26 93
1
=
=
=
=
∠ =
H
H
H
H
H
m
29 = 5 29
≈ . ft
9. α 880 39 44
180 83 97
180 97 83
º º º
º º º
º º º
− +( ) =
− =
∠ = − =10. m β
ssupplementary angles
isosc
( )11. right triangle, eeles
major arc and minor arc add up
to 360º:
12.
3360 79 281
120
º º º
º
− =
− =
13. exterior angles are
180º 660
36060
6
º
ºº
exterior angles add up to 360º:
= sidd
bh
es: hexagon
area of one triangle14. =
=12
12
7 5.(( ) ( ) =
( ) =
5 18 75 2
112
.
.
in
six triangles:
6 18.75 in2 55 2
6 7 5 45.
in
P in in
rHsB
15.
16.
= ( ) =
is a rhombus
117.
18.
Bs definition of a rhombus
def
≅
≅
rH
Br Hs
:
: iinition of a rhombus
reflexive proper19. BH BH≅ : tty
HsB sss20. ≅ :BrH
sYsteMatic reVieW 28D - Lesson Practice 29a
soLUtions GeoMetrY200
Systematic Review 28D1.
2.
3.
4.
5.
6. B
B
15 646
6 4
+( ) =
= 115 6
6 4 21
6 84
14
15 104
10
10 4 1
+( )= ( )=
=
+( ) =
=
B
B
B
B
B
ft
7.
55 10
10 4 25
10 100
10
102 252 2
100 6
+( )= ( )=
=
+ =
+
B
B
B
H
ft
8.
225 2
725 2
725
25
26 93
1
=
=
=
=
∠ =
H
H
H
H
H
m
29 = 5 29
≈ . ft
9. α 880 39 44
180 83 97
180 97 83
º º º
º º º
º º º
− +( ) =
− =
∠ = − =10. m β
ssupplementary angles
isosc
( )11. right triangle, eeles
major arc and minor arc add up
to 360º:
12.
3360 79 281
120
º º º
º
− =
− =
13. exterior angles are
180º 660
36060
6
º
ºº
exterior angles add up to 360º:
= sidd
bh
es: hexagon
area of one triangle14. =
=12
12
7 5.(( ) ( ) =
( ) =
5 18 75 2
112
.
.
in
six triangles:
6 18.75 in2 55 2
6 7 5 45.
in
P in in
rHsB
15.
16.
= ( ) =
is a rhombus
117.
18.
Bs definition of a rhombus
def
≅
≅
rH
Br Hs
:
: iinition of a rhombus
reflexive proper19. BH BH≅ : tty
HsB sss20. ≅ :BrH
Systematic Review 28D
Systematic Review 28ESystematic Review 28E1- 4.
5.
6. X
X
X
15020
20 50
502
=
=
=00
2 12
2 5
502 2 5
=
+
. yd
.
or
(or 7.5 ft)
7. 22 2
2 500 6 25 2
2 506 25 2
2 506 25 50 06
=
+ =
=
=
H
H
H
H y
, .
, .
, . .≈ dd (or 150.2 ft)
exterior angle8. = − =180 108 72
3
º º º
66072
5
12
ºº
:=
=
pentagon
of one triangle:9. area
a bhh m= ( ) ( ) =12
11 8 44 2
5
area of five triangles:
44 m2 (( ) =
= ( ) =
∠
220 2
5 11 55
m
P m m
KZ
10.
11. bisects JKX annd JZX
numbers 12-14 may be in any order.)
∠
(
12. ∠∠ ≅ ∠
∠ ≅ ∠
JKZ XKZ
JZK XZK
:
:
definition of bisector
13.
definition of bisector
KZ reflexive pro14. ≅ KZ : pperty
JKZ asa15.
16.
≅
= +( ) + +( )
+(
:XKZ
P X X
X
2 3 2 1
2 3)) + +( ) =
+ + + =+ =
==
=
2 1 36
2 6 2 2 364 8 36
4 287
X
X XX
XX
a
ft
17. bbh X X
X X
= +( ) ( )
+( ) ( ) => +( ) ( ) = ( ) ( ) =
3
3 7 3 7 10 7 70 2ft
18.. P Y Y
Y Y
Y YY
= +( ) + −( )
+( ) + −( ) =
+ + − =
2 4 2 1
2 4 2 1 98
2 8 2 2 984 ++ =
==
= = +( ) −( )+( ) −( ) =
6 984 92
23
4 1
4 1
YY in
a bh Y Y
Y Y
19.
>>
( ) +( ) ( ) −( ) =
( )( ) =
+ =
23 4 23 1
27 22 594 2
5 4 9º º
in
X X20. 00
9 90
10
º:
º º
º
complementary angles
X
X
=
=
2
4
Systematic Review 28E1- 4.
5.
6. X
X
X
15020
20 50
502
=
=
=00
2 12
2 5
502 2 5
=
+
. yd
.
or
(or 7.5 ft)
7. 22 2
2 500 6 25 2
2 506 25 2
2 506 25 50 06
=
+ =
=
=
H
H
H
H y
, .
, .
, . .≈ dd (or 150.2 ft)
exterior angle8. = − =180 108 72
3
º º º
66072
5
12
ºº
:=
=
pentagon
of one triangle:9. area
a bhh m= ( ) ( ) =12
11 8 44 2
5
area of five triangles:
44 m2 (( ) =
= ( ) =
∠
220 2
5 11 55
m
P m m
KZ
10.
11. bisects JKX annd JZX
numbers 12-14 may be in any order.)
∠
(
12. ∠∠ ≅ ∠
∠ ≅ ∠
JKZ XKZ
JZK XZK
:
:
definition of bisector
13.
definition of bisector
KZ reflexive pro14. ≅ KZ : pperty
JKZ asa15.
16.
≅
= +( ) + +( )
+(
:XKZ
P X X
X
2 3 2 1
2 3)) + +( ) =
+ + + =+ =
==
=
2 1 36
2 6 2 2 364 8 36
4 287
X
X XX
XX
a
ft
17. bbh X X
X X
= +( ) ( )
+( ) ( ) => +( ) ( ) = ( ) ( ) =
3
3 7 3 7 10 7 70 2ft
18.. P Y Y
Y Y
Y YY
= +( ) + −( )
+( ) + −( ) =
+ + − =
2 4 2 1
2 4 2 1 98
2 8 2 2 984 ++ =
==
= = +( ) −( )+( ) −( ) =
6 984 92
23
4 1
4 1
YY in
a bh Y Y
Y Y
19.
>>
( ) +( ) ( ) −( ) =
( )( ) =
+ =
23 4 23 1
27 22 594 2
5 4 9º º
in
X X20. 00
9 90
10
º:
º º
º
complementary angles
X
X
=
=
20 yd
X 120
30 1 X
3 ft = 1 yd
3 ft = 1 yd Systematic Review 28E1- 4.
5.
6. X
X
X
15020
20 50
502
=
=
=00
2 12
2 5
502 2 5
=
+
. yd
.
or
(or 7.5 ft)
7. 22 2
2 500 6 25 2
2 506 25 2
2 506 25 50 06
=
+ =
=
=
H
H
H
H y
, .
, .
, . .≈ dd (or 150.2 ft)
exterior angle8. = − =180 108 72
3
º º º
66072
5
12
ºº
:=
=
pentagon
of one triangle:9. area
a bhh m= ( ) ( ) =12
11 8 44 2
5
area of five triangles:
44 m2 (( ) =
= ( ) =
∠
220 2
5 11 55
m
P m m
KZ
10.
11. bisects JKX annd JZX
numbers 12-14 may be in any order.)
∠
(
12. ∠∠ ≅ ∠
∠ ≅ ∠
JKZ XKZ
JZK XZK
:
:
definition of bisector
13.
definition of bisector
KZ reflexive pro14. ≅ KZ : pperty
JKZ asa15.
16.
≅
= +( ) + +( )
+(
:XKZ
P X X
X
2 3 2 1
2 3)) + +( ) =
+ + + =+ =
==
=
2 1 36
2 6 2 2 364 8 36
4 287
X
X XX
XX
a
ft
17. bbh X X
X X
= +( ) ( )
+( ) ( ) => +( ) ( ) = ( ) ( ) =
3
3 7 3 7 10 7 70 2ft
18.. P Y Y
Y Y
Y YY
= +( ) + −( )
+( ) + −( ) =
+ + − =
2 4 2 1
2 4 2 1 98
2 8 2 2 984 ++ =
==
= = +( ) −( )+( ) −( ) =
6 984 92
23
4 1
4 1
YY in
a bh Y Y
Y Y
19.
>>
( ) +( ) ( ) −( ) =
( )( ) =
+ =
23 4 23 1
27 22 594 2
5 4 9º º
in
X X20. 00
9 90
10
º:
º º
º
complementary angles
X
X
=
=
Systematic Review 28E1- 4.
5.
6. X
X
X
15020
20 50
502
=
=
=00
2 12
2 5
502 2 5
=
+
. yd
.
or
(or 7.5 ft)
7. 22 2
2 500 6 25 2
2 506 25 2
2 506 25 50 06
=
+ =
=
=
H
H
H
H y
, .
, .
, . .≈ dd (or 150.2 ft)
exterior angle8. = − =180 108 72
3
º º º
66072
5
12
ºº
:=
=
pentagon
of one triangle:9. area
a bhh m= ( ) ( ) =12
11 8 44 2
5
area of five triangles:
44 m2 (( ) =
= ( ) =
∠
220 2
5 11 55
m
P m m
KZ
10.
11. bisects JKX annd JZX
numbers 12-14 may be in any order.)
∠
(
12. ∠∠ ≅ ∠
∠ ≅ ∠
JKZ XKZ
JZK XZK
:
:
definition of bisector
13.
definition of bisector
KZ reflexive pro14. ≅ KZ : pperty
JKZ asa15.
16.
≅
= +( ) + +( )
+(
:XKZ
P X X
X
2 3 2 1
2 3)) + +( ) =
+ + + =+ =
==
=
2 1 36
2 6 2 2 364 8 36
4 287
X
X XX
XX
a
ft
17. bbh X X
X X
= +( ) ( )
+( ) ( ) => +( ) ( ) = ( ) ( ) =
3
3 7 3 7 10 7 70 2ft
18.. P Y Y
Y Y
Y YY
= +( ) + −( )
+( ) + −( ) =
+ + − =
2 4 2 1
2 4 2 1 98
2 8 2 2 984 ++ =
==
= = +( ) −( )+( ) −( ) =
6 984 92
23
4 1
4 1
YY in
a bh Y Y
Y Y
19.
>>
( ) +( ) ( ) −( ) =
( )( ) =
+ =
23 4 23 1
27 22 594 2
5 4 9º º
in
X X20. 00
9 90
10
º:
º º
º
complementary angles
X
X
=
=
Lesson Practice 29ALesson Practice 29A1.
2.
3.
4.
5.
6
51312135
1212135
13
..
7.
8.
9.
10.
11.
12.
13.
125725242572424257252476
18= 11
317186
1717186
1813
1768
1045
14.
15.
16.
17.
18.
19.
=
=
220.
21.
22.
23.
24.
610
35
86
43
610
35
810
45
68
34
=
=
=
=
=
Lesson Practice 29a - Lesson Practice 29B
soLUtionsGeoMetrY 201
Lesson Practice 29A1.
2.
3.
4.
5.
6
51312135
1212135
13
..
7.
8.
9.
10.
11.
12.
13.
125725242572424257252476
18= 11
317186
1717186
1813
1768
1045
14.
15.
16.
17.
18.
19.
=
=
220.
21.
22.
23.
24.
610
35
86
43
610
35
810
45
68
34
=
=
=
=
=
Lesson Practice 29BLesson Practice 29B1.
2.
5
5 2
1
2
2
2
22
5
5 2
1
2
2
2
22
= × =
= × =
33.
4.
5.
6.
7.
55
1
5
5 2
1
2
2
2
22
5
5 2
1
2
2
2
22
55
1
1326
=
= × =
= × =
=
==
=
= × =
=
12
13 326
32
13
13 3
1
3
3
3
33
13 326
32
132
8.
9.
10.
11.66
12
13 313
31
3
7
130
130
130
7 130130
9
130
=
= =
× =
12.
13.
14. ×× =
× =
130
130
9 130130
79
9
130
130
130
9 130130
7
15.
16.
17.1130
130
130
7 130130
97
20318
1118
20311
× =
18.
19.
20.
21.
222.
23.
24.
1118
20318
11
203
203
203
11 203203
× =
Lesson Practice 29B1.
2.
5
5 2
1
2
2
2
22
5
5 2
1
2
2
2
22
= × =
= × =
33.
4.
5.
6.
7.
55
1
5
5 2
1
2
2
2
22
5
5 2
1
2
2
2
22
55
1
1326
=
= × =
= × =
=
==
=
= × =
=
12
13 326
32
13
13 3
1
3
3
3
33
13 326
32
132
8.
9.
10.
11.66
12
13 313
31
3
7
130
130
130
7 130130
9
130
=
= =
× =
12.
13.
14. ×× =
× =
130
130
9 130130
79
9
130
130
130
9 130130
7
15.
16.
17.1130
130
130
7 130130
97
20318
1118
20311
× =
18.
19.
20.
21.
222.
23.
24.
1118
20318
11
203
203
203
11 203203
× =
sYsteMatic reVieW 29c - sYsteMatic reVieW 29D
soLUtions GeoMetrY202
Systematic Review 29CSystematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
Systematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
Systematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
16 14
Systematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
2 ft
2 ft
10 ft
X
X
10 ft 15 ft 2 ft 2 ft
Systematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
Systematic Review 29C1.
2.
3.
3050
35
4050
45
3040
3
=
=
=44
4050
45
3050
35
4030
43
4.
5.
6.
7.
8.
9.
10.
1
=
=
=
acBcaBBc
11.
12.
13 -16.
17.
18.
19.
acBa
X
X
X
X
252
1010 50
5
10
=
=
= ft
002
1010 200
20
42
1010 8
810
45
=
=
=
=
=
= =
X
X
X
X
X
ft
ft
20.
Systematic Review 29DSystematic Review 29D1.
2.
3 512 5
1010
35125
725
..
× = =
11212 5
1010
120125
2425
3 512
1010
35120
724
..
× = =
= = =3.
4..
5.
1212 5
1010
120125
2425
3 512 5
1010
35125
7...
× = =
× = =225
123 5
1010
12035
247
498 59
1010
8590
1
6.
7.
8.
.
.
× = =
× = = 7718
48 5
1010
4085
817
8 59
1010
8590
1718
9.
10.
1
.
.
× = =
× = =
11.
12.
13 -16.
17.
498 54
1010
8540
178
. × = =
Q is the midppoint of Ut given
definition of midp
:
:18. UQ tQ≅ ooint
vertical angles
QsU
19.
20.
∠ ≅ ∠
≅
UQs tQP
QP
:
tt : aas
14
16
Systematic Review 29D1.
2.
3 512 5
1010
35125
725
..
× = =
11212 5
1010
120125
2425
3 512
1010
35120
724
..
× = =
= = =3.
4..
5.
1212 5
1010
120125
2425
3 512 5
1010
35125
7...
× = =
× = =225
123 5
1010
12035
247
498 59
1010
8590
1
6.
7.
8.
.
.
× = =
× = = 7718
48 5
1010
4085
817
8 59
1010
8590
1718
9.
10.
1
.
.
× = =
× = =
11.
12.
13 -16.
17.
498 54
1010
8540
178
. × = =
Q is the midppoint of Ut given
definition of midp
:
:18. UQ tQ≅ ooint
vertical angles
QsU
19.
20.
∠ ≅ ∠
≅
UQs tQP
QP
:
tt : aas
sYsteMatic reVieW 29D - Lesson Practice 30a
soLUtionsGeoMetrY 203
Systematic Review 29D1.
2.
3 512 5
1010
35125
725
..
× = =
11212 5
1010
120125
2425
3 512
1010
35120
724
..
× = =
= = =3.
4..
5.
1212 5
1010
120125
2425
3 512 5
1010
35125
7...
× = =
× = =225
123 5
1010
12035
247
498 59
1010
8590
1
6.
7.
8.
.
.
× = =
× = = 7718
48 5
1010
4085
817
8 59
1010
8590
1718
9.
10.
1
.
.
× = =
× = =
11.
12.
13 -16.
17.
498 54
1010
8540
178
. × = =
Q is the midppoint of Ut given
definition of midp
:
:18. UQ tQ≅ ooint
vertical angles
QsU
19.
20.
∠ ≅ ∠
≅
UQs tQP
QP
:
tt : aas
Systematic Review 29D
Systematic Review 29ESystematic Review 29E1.
2.
812 8
1010
80128
58
101
.× = =
22 81010
100128
2532
810
45
1012 8
1010
1001
.
.
× = =
=
× =
3.
4.228
2532
812 8
1010
80128
58
108
54
=
× = =
=
5.
6.
7 -10.
10.
.
apppears the same as the original,
because the figuure is symmetrical
around the X axis.
equiang11. uular triangle, equilateral
long leg ÷ 312. = 5 6
3== =
× = ( )( ) =
5 21
5 2
2 5 2 2 10 213.
14.
15
short
diameter
leg
..
16.
17.
18.
40 85 125
180 125 55
180
180 72
º º º
º º º
º
º
+ =
− =
− ºº º
º
=
( )+ =
108
2 3 180
supplementary angles
sup
19. B B
pplementary angles
alternate
( )=
=
=
5 180
36
3
B
B
B X
º
º
iinterior angles( )= ( ) =
=
+ =
X
X
B B
3 36 108
108
6 14
º º
º
20. 1180
20 180
9
6
º
º
º
supplementary angles
al
( )=
=
=
B
B
B X tternate exterior angles( )= ( ) =X 6 9 54º º
8
10
Systematic Review 29E1.
2.
812 8
1010
80128
58
101
.× = =
22 81010
100128
2532
810
45
1012 8
1010
1001
.
.
× = =
=
× =
3.
4.228
2532
812 8
1010
80128
58
108
54
=
× = =
=
5.
6.
7 -10.
10.
.
apppears the same as the original,
because the figuure is symmetrical
around the X axis.
equiang11. uular triangle, equilateral
long leg ÷ 312. = 5 6
3== =
× = ( )( ) =
5 21
5 2
2 5 2 2 10 213.
14.
15
short
diameter
leg
..
16.
17.
18.
40 85 125
180 125 55
180
180 72
º º º
º º º
º
º
+ =
− =
− ºº º
º
=
( )+ =
108
2 3 180
supplementary angles
sup
19. B B
pplementary angles
alternate
( )=
=
=
5 180
36
3
B
B
B X
º
º
iinterior angles( )= ( ) =
=
+ =
X
X
B B
3 36 108
108
6 14
º º
º
20. 1180
20 180
9
6
º
º
º
supplementary angles
al
( )=
=
=
B
B
B X tternate exterior angles( )= ( ) =X 6 9 54º º
Systematic Review 29E1.
2.
812 8
1010
80128
58
101
.× = =
22 81010
100128
2532
810
45
1012 8
1010
1001
.
.
× = =
=
× =
3.
4.228
2532
812 8
1010
80128
58
108
54
=
× = =
=
5.
6.
7 -10.
10.
.
apppears the same as the original,
because the figuure is symmetrical
around the X axis.
equiang11. uular triangle, equilateral
long leg ÷ 312. = 5 6
3== =
× = ( )( ) =
5 21
5 2
2 5 2 2 10 213.
14.
15
short
diameter
leg
..
16.
17.
18.
40 85 125
180 125 55
180
180 72
º º º
º º º
º
º
+ =
− =
− ºº º
º
=
( )+ =
108
2 3 180
supplementary angles
sup
19. B B
pplementary angles
alternate
( )=
=
=
5 180
36
3
B
B
B X
º
º
iinterior angles( )= ( ) =
=
+ =
X
X
B B
3 36 108
108
6 14
º º
º
20. 1180
20 180
9
6
º
º
º
supplementary angles
al
( )=
=
=
B
B
B X tternate exterior angles( )= ( ) =X 6 9 54º º
Lesson Practice 30ALesson Practice 30A1.
2.
3.
4.
5
610
35
810
45
68
34
53
=
=
=
..
6.
7.
8.
9.
10.
1
54432426
1213
1026
513
2410
125
1312
=
=
=
11.
12.
13.
14.
1355
124
8 51010
4085
817
7 58 5
1010
.
.
.
× = =
× == =
× = =
7585
1517
47 5
1010
4075
815
178
1715
15.
16.
17.
18
.
..
19.
20.
158
1
Pythagorean theorem
Lesson Practice 30a - sYsteMatic reVieW 30D
soLUtions GeoMetrY204
Lesson Practice 30A1.
2.
3.
4.
5
610
35
810
45
68
34
53
=
=
=
..
6.
7.
8.
9.
10.
1
54432426
1213
1026
513
2410
125
1312
=
=
=
11.
12.
13.
14.
1355
124
8 51010
4085
817
7 58 5
1010
.
.
.
× = =
× == =
× = =
7585
1517
47 5
1010
4075
815
178
1715
15.
16.
17.
18
.
..
19.
20.
158
1
Pythagorean theorem
Lesson Practice 30BLesson Practice 30B1.
2.
3.
4.
5.
6.
7.
354534535443151178
17158171517881548
12
5 78
8.
9.
10.
11.
12.
13.
14.
=
. ×× =
× =
1010
5780
45 7
1010
4057
2180575740
15.
16.
17.
18.
.
119.
20.
cos2θ
trig ratios
Systematic Review 30CSystematic Review 30C1.
2
1213 4
1010
120134
6067.
× = =
..
3.
4.
5.
6.
613 4
1010
60134
3067
126
21
6760673012
.× = =
=
77.
8 -11.
12.
sin cos2 2 1θ θ+ =
∠ ≅ ∠LPG rGP:
definition off parallelogram;
alternate interior angles
13. ∠LGGP rPG≅ ∠ :
definition of parallelogram;
alternate interior angles
PG reflexive property14.
15
≅ PG :
..
16.
17.
18.
PLG
decagon
congruent
theor
≅ :GrP asa
eems
arc
2
19.
20.
Systematic Review 30C1.
2
1213 4
1010
120134
6067.
× = =
..
3.
4.
5.
6.
613 4
1010
60134
3067
126
21
6760673012
.× = =
=
77.
8 -11.
12.
sin cos2 2 1θ θ+ =
∠ ≅ ∠LPG rGP:
definition off parallelogram;
alternate interior angles
13. ∠LGGP rPG≅ ∠ :
definition of parallelogram;
alternate interior angles
PG reflexive property14.
15
≅ PG :
..
16.
17.
18.
PLG
decagon
congruent
theor
≅ :GrP asa
eems
arc
2
19.
20.
9
11
X = –1
Systematic Review 30C1.
2
1213 4
1010
120134
6067.
× = =
..
3.
4.
5.
6.
613 4
1010
60134
3067
126
21
6760673012
.× = =
=
77.
8 -11.
12.
sin cos2 2 1θ θ+ =
∠ ≅ ∠LPG rGP:
definition off parallelogram;
alternate interior angles
13. ∠LGGP rPG≅ ∠ :
definition of parallelogram;
alternate interior angles
PG reflexive property14.
15
≅ PG :
..
16.
17.
18.
PLG
decagon
congruent
theor
≅ :GrP asa
eems
arc
2
19.
20.
Systematic Review 30DSystematic Review 30D1.
2
13 615
1010
136150
6875
. × = =
..
3.
6 415
1010
64150
3275
13 66 4
1010
13664
178
.
..
× = =
× = =
44.
5.
1513 6
1010
150136
7568
7532
.× = =
(reciprocal of #2)
(reciprocal of #3)6.
7.
8 - 9.
10.
8171
ae cD|| : ggiven
Bea
alternate interior angles
11.
12
∠ ≅ ∠BcD :
..
13.
14.
∠ ≅ ∠aBe vertical anglesDBc
aBe DBc aa
:
~ :
cchord
tangent
3
15.
16.
17.
18.
19.
180
360
13
13
6
º
º
V Bh= = 00 60 50
60 000 3
4 12
60 70
×( ) ( ) =
= ( ) ( )
,
[ ]
mm
sa20. ++ ( ) ( ) =
+ =
60 60
8 400 3 600 12 000 2, , , mm
sYsteMatic reVieW 30D - sYsteMatic reVieW 30e
soLUtionsGeoMetrY 205
Systematic Review 30D1.
2
13 615
1010
136150
6875
. × = =
..
3.
6 415
1010
64150
3275
13 66 4
1010
13664
178
.
..
× = =
× = =
44.
5.
1513 6
1010
150136
7568
7532
.× = =
(reciprocal of #2)
(reciprocal of #3)6.
7.
8 - 9.
10.
8171
ae cD|| : ggiven
Bea
alternate interior angles
11.
12
∠ ≅ ∠BcD :
..
13.
14.
∠ ≅ ∠aBe vertical anglesDBc
aBe DBc aa
:
~ :
cchord
tangent
3
15.
16.
17.
18.
19.
180
360
13
13
6
º
º
V Bh= = 00 60 50
60 000 3
4 12
60 70
×( ) ( ) =
= ( ) ( )
,
[ ]
mm
sa20. ++ ( ) ( ) =
+ =
60 60
8 400 3 600 12 000 2, , , mm
X = 1
9
Systematic Review 30D1.
2
13 615
1010
136150
6875
. × = =
..
3.
6 415
1010
64150
3275
13 66 4
1010
13664
178
.
..
× = =
× = =
44.
5.
1513 6
1010
150136
7568
7532
.× = =
(reciprocal of #2)
(reciprocal of #3)6.
7.
8 - 9.
10.
8171
ae cD|| : ggiven
Bea
alternate interior angles
11.
12
∠ ≅ ∠BcD :
..
13.
14.
∠ ≅ ∠aBe vertical anglesDBc
aBe DBc aa
:
~ :
cchord
tangent
3
15.
16.
17.
18.
19.
180
360
13
13
6
º
º
V Bh= = 00 60 50
60 000 3
4 12
60 70
×( ) ( ) =
= ( ) ( )
,
[ ]
mm
sa20. ++ ( ) ( ) =
+ =
60 60
8 400 3 600 12 000 2, , , mm
Systematic Review 30ESystematic Review 30E1.
2.
3.
4.
5.
242572524725242577724
6.
7.
8 - 9.
10.
11.
12.
sin
octagon
perpendicular
θ
rr d
a r
= = ( ) =
= ( ) ( ) =
12
12
20 10
2 3 14 102
314 2
ft
. ftπ ≈
13. eexpress all measurements
in the same unit:
.75 fft 12× =
× =
= = ( ) ( ) × =
9
1 5 12 18
12
9 5 18 405
. ft
in
in
V Bh in
sa
3
14. = area of 2 ends plus area of
2 sides pplus area of bottom =
( ) ( ) + ( ) ( )2 12
9 5 2 18 7( ) (( ) + ( ) ( ) =
+ + =
= ( )
9 18
45 252 162 459 2
2 2 3 14 4.
in
D r15. π ≈ (( ) =
=
×
25 12
14
25
. in
of16. arc is 90º360º
circle
14
.. .12 6 28
3 72
14
2 102
14
2 10 2 1
=
+( ) + +( )=
+ =
+ =
in
X X
X
X
17.
44
2 10 282 18
9
( )+ =
==
= ×
XXX in
a average height18. base
= +( )
+( ) => ( ) +( ) = ( ) =
14 2
14 2 14 9 2 14 11 154 2
X
X in
199.
20.
36 2 3 2 3 2
36 6 6
30 6
5
42
= +( ) + +( ) + ( )= +
=
=
a a a
a
a
a ft
== +( ) + +( ) + ( )= +
=
=
2 3 2 3 2
42 6 6
36 6
6
a a a
a
a
a ft
Systematic Review 30E1.
2.
3.
4.
5.
242572524725242577724
6.
7.
8 - 9.
10.
11.
12.
sin
octagon
perpendicular
θ
rr d
a r
= = ( ) =
= ( ) ( ) =
12
12
20 10
2 3 14 102
314 2
ft
. ftπ ≈
13. eexpress all measurements
in the same unit:
.75 fft 12× =
× =
= = ( ) ( ) × =
9
1 5 12 18
12
9 5 18 405
. ft
in
in
V Bh in
sa
3
14. = area of 2 ends plus area of
2 sides pplus area of bottom =
( ) ( ) + ( ) ( )2 12
9 5 2 18 7( ) (( ) + ( ) ( ) =
+ + =
= ( )
9 18
45 252 162 459 2
2 2 3 14 4.
in
D r15. π ≈ (( ) =
=
×
25 12
14
25
. in
of16. arc is 90º360º
circle
14
.. .12 6 28
3 72
14
2 102
14
2 10 2 1
=
+( ) + +( )=
+ =
+ =
in
X X
X
X
17.
44
2 10 282 18
9
( )+ =
==
= ×
XXX in
a average height18. base
= +( )
+( ) => ( ) +( ) = ( ) =
14 2
14 2 14 9 2 14 11 154 2
X
X in
199.
20.
36 2 3 2 3 2
36 6 6
30 6
5
42
= +( ) + +( ) + ( )= +
=
=
a a a
a
a
a ft
== +( ) + +( ) + ( )= +
=
=
2 3 2 3 2
42 6 6
36 6
6
a a a
a
a
a ft
Systematic Review 30E1.
2.
3.
4.
5.
242572524725242577724
6.
7.
8 - 9.
10.
11.
12.
sin
octagon
perpendicular
θ
rr d
a r
= = ( ) =
= ( ) ( ) =
12
12
20 10
2 3 14 102
314 2
ft
. ftπ ≈
13. eexpress all measurements
in the same unit:
.75 fft 12× =
× =
= = ( ) ( ) × =
9
1 5 12 18
12
9 5 18 405
. ft
in
in
V Bh in
sa
3
14. = area of 2 ends plus area of
2 sides pplus area of bottom =
( ) ( ) + ( ) ( )2 12
9 5 2 18 7( ) (( ) + ( ) ( ) =
+ + =
= ( )
9 18
45 252 162 459 2
2 2 3 14 4.
in
D r15. π ≈ (( ) =
=
×
25 12
14
25
. in
of16. arc is 90º360º
circle
14
.. .12 6 28
3 72
14
2 102
14
2 10 2 1
=
+( ) + +( )=
+ =
+ =
in
X X
X
X
17.
44
2 10 282 18
9
( )+ =
==
= ×
XXX in
a average height18. base
= +( )
+( ) => ( ) +( ) = ( ) =
14 2
14 2 14 9 2 14 11 154 2
X
X in
199.
20.
36 2 3 2 3 2
36 6 6
30 6
5
42
= +( ) + +( ) + ( )= +
=
=
a a a
a
a
a ft
== +( ) + +( ) + ( )= +
=
=
2 3 2 3 2
42 6 6
36 6
6
a a a
a
a
a ft
sYsteMatic reVieW 30e - sYsteMatic reVieW 30e
soLUtions GeoMetrY206
Systematic Review 30E1.
2.
3.
4.
5.
242572524725242577724
6.
7.
8 - 9.
10.
11.
12.
sin
octagon
perpendicular
θ
rr d
a r
= = ( ) =
= ( ) ( ) =
12
12
20 10
2 3 14 102
314 2
ft
. ftπ ≈
13. eexpress all measurements
in the same unit:
.75 fft 12× =
× =
= = ( ) ( ) × =
9
1 5 12 18
12
9 5 18 405
. ft
in
in
V Bh in
sa
3
14. = area of 2 ends plus area of
2 sides pplus area of bottom =
( ) ( ) + ( ) ( )2 12
9 5 2 18 7( ) (( ) + ( ) ( ) =
+ + =
= ( )
9 18
45 252 162 459 2
2 2 3 14 4.
in
D r15. π ≈ (( ) =
=
×
25 12
14
25
. in
of16. arc is 90º360º
circle
14
.. .12 6 28
3 72
14
2 102
14
2 10 2 1
=
+( ) + +( )=
+ =
+ =
in
X X
X
X
17.
44
2 10 282 18
9
( )+ =
==
= ×
XXX in
a average height18. base
= +( )
+( ) => ( ) +( ) = ( ) =
14 2
14 2 14 9 2 14 11 154 2
X
X in
199.
20.
36 2 3 2 3 2
36 6 6
30 6
5
42
= +( ) + +( ) + ( )= +
=
=
a a a
a
a
a ft
== +( ) + +( ) + ( )= +
=
=
2 3 2 3 2
42 6 6
36 6
6
a a a
a
a
a ft
Honors Lesson 1 - Honors Lesson 3GeoMetrY soLUtions 207
Honors Lesson 1Lesson11. Begin by putting x's to show that tyleer
and Madison do not like tacos. that
leaves JJeff as the one who has tacos as
his favorite. since you know Jeff's
favorite, you can also pput x's in Jeff's
row, under ice cream and steaak. We are
told that Madison is allergic to anyything
made with milk, so we can put an x
acrosss from her name, under ice cream.
now we can ssee that tyler is the only one
who can have icee cream as his favorite,
leaving Madison with ssteak.
cream tacos steakX
X
iceJeff X yes
tyler yes X
Maddison X X
We
yes
use similar reasoning for the r2. eest of
the problems. remember that once you
havve a "yes" in any row or column, the
rest of thhe possibilities in that row and
in that columnn can be eliminated.
brown blondeX
blackMike yes X
caaitlyn X X
Lisa X yes
reading tennis
yes
X
cooking eati3. nng
X
X
yes
George X X yes X
celia X yes X
Donna yes X X
adam X X X
44. spring autumn wintersummer
David X X yes X
Linda X X X yees
shauna yes X X
april X yes X X
X
Honors Lesson 2Lesson 21. 18 20 38
38 30 8
+ =
− = days had both
sun rain
100 8 12
s r = 8
s r = 30
1 (the twisted
∩
∪
2. rring you started with
is called a Mobius strip.))
1st time : one long loop is created
2nd time
3.
: two interlocked loops
are created
4. 52 5 6+( ) ÷ + =
+( ) + =
( ) + =
+ =
10
25 5 6 10
30 6 10
5 10 15
42
÷
÷
÷5. 77 6 1
6 6 1
12 1 11
( ) + − =
( ) + − =
− =
Honors Lesson 3Lesson 31.
2.
3.
4.
5.
6.
7.
8.
2
4
3
12
8
3
12
L F
L F
K M P
∩ =
∪ =
∪ − =
KK M P
x
∩ ∩ =
=
4
3 4 49. x 3
commutative property
is true ffor multiplication
commutative proper
10. 9 6 6 9− −≠
tty
is false for subtraction
as
11. 2 1 5 2 1 5+( ) + = + +( )ssociative property
is true for addition
12. 2 8 8÷ ≠ ÷÷2
commutative property
is false for division
7
2
1
2
3
4 seeds
leaves
flowers
Lesson 31.
2.
3.
4.
5.
6.
7.
8.
2
4
3
12
8
3
12
L F
L F
K M P
∩ =
∪ =
∪ − =
KK M P
x
∩ ∩ =
=
4
3 4 49. x 3
commutative property
is true ffor multiplication
commutative proper
10. 9 6 6 9− −≠
tty
is false for subtraction
as
11. 2 1 5 2 1 5+( ) + = + +( )ssociative property
is true for addition
12. 2 8 8÷ ≠ ÷÷2
commutative property
is false for division
Honors Solutions
Honors soLUtions
Honors Lesson 3 - Honors Lesson 5
soLUtions GeoMetrY208
Lesson 31.
2.
3.
4.
5.
6.
7.
8.
2
4
3
12
8
3
12
L F
L F
K M P
∩ =
∪ =
∪ − =
KK M P
x
∩ ∩ =
=
4
3 4 49. x 3
commutative property
is true ffor multiplication
commutative proper
10. 9 6 6 9− −≠
tty
is false for subtraction
as
11. 2 1 5 2 1 5+( ) + = + +( )ssociative property
is true for addition
12. 2 8 8÷ ≠ ÷÷2
commutative property
is false for division
1
0 4
2
3
1
8 pickles
mustard
ketchup
Lesson 31.
2.
3.
4.
5.
6.
7.
8.
2
4
3
12
8
3
12
L F
L F
K M P
∩ =
∪ =
∪ − =
KK M P
x
∩ ∩ =
=
4
3 4 49. x 3
commutative property
is true ffor multiplication
commutative proper
10. 9 6 6 9− −≠
tty
is false for subtraction
as
11. 2 1 5 2 1 5+( ) + = + +( )ssociative property
is true for addition
12. 2 8 8÷ ≠ ÷÷2
commutative property
is false for division
Honors Lesson 4Lesson 41.
2.
3.
4.
45º
nnW
nne
no, he should have corrrected 67.5º
5.
6.
5 6 2 18
5 2 18 6
3 24
8
2
X X
X X
X
X
c
− = +
− = +
=
=
++ = −
=
=
+( ) + =
+ =
10 43
3 33
11
1 75 3 25
2 1 75
c
c
c
D D
D
7. $ . $ .
$ . $33 25
2 1 50
75
75 1 75 2 50
.
$ .
$.
$. $ . $ .
D
D
=
=
+ =
Drink is $.755
sandwich is $2.50
let X = number of isaac's 8. ccustomers
2X = number of aaron's customers
X+2X == 105
3X = 105
X = 35
2X = 70
isaac has 35 customerrs
aaron has 70 customers
feet
9. X X
X
X
+ =
=
=
2 18
3 18
6 ;; 2X = 12 feet
10. a a
a
a
a
+ +( ) =
+ =
=
=
20 144
2 20 144
2 124
662 apples in one box
62 + 20 = 82 apples in the other box
Lesson 41.
2.
3.
4.
45º
nnW
nne
no, he should have corrrected 67.5º
5.
6.
5 6 2 18
5 2 18 6
3 24
8
2
X X
X X
X
X
c
− = +
− = +
=
=
++ = −
=
=
+( ) + =
+ =
10 43
3 33
11
1 75 3 25
2 1 75
c
c
c
D D
D
7. $ . $ .
$ . $33 25
2 1 50
75
75 1 75 2 50
.
$ .
$.
$. $ . $ .
D
D
=
=
+ =
Drink is $.755
sandwich is $2.50
let X = number of isaac's 8. ccustomers
2X = number of aaron's customers
X+2X == 105
3X = 105
X = 35
2X = 70
isaac has 35 customerrs
aaron has 70 customers
feet
9. X X
X
X
+ =
=
=
2 18
3 18
6 ;; 2X = 12 feet
10. a a
a
a
a
+ +( ) =
+ =
=
=
20 144
2 20 144
2 124
662 apples in one box
62 + 20 = 82 apples in the other box
Honors Lesson 5Lesson 5
1.
2.
3.
4 small 2 medium
1 large
7 total
44. 1 started with
2 that are half of first trianggle
6 small
7 overlapping (you may need to
16 tottal draw these
separately to be
able to count each
one. see above.)
5.
6.
Lesson 51.
2.
3.
4 small 2 medium
1 large
7 total
44. 1 started with
2 that are half of first trianggle
6 small
7 overlapping (you may need to
16 tottal draw these
separately to be
able to count each
one. see above.)
5.
6.
Lesson 51.
2.
3.
4 small 2 medium
1 large
7 total
44. 1 started with
2 that are half of first trianggle
6 small
7 overlapping (you may need to
16 tottal draw these
separately to be
able to count each
one. see above.)
5.
6.
Honors Lesson 5 - Honors Lesson 7
soLUtionsGeoMetrY 209
Lesson 51.
2.
3.
4 small 2 medium
1 large
7 total
44. 1 started with
2 that are half of first trianggle
6 small
7 overlapping (you may need to
16 tottal draw these
separately to be
able to count each
one. see above.)
5.
6.
Lesson 51.
2.
3.
4 small 2 medium
1 large
7 total
44. 1 started with
2 that are half of first trianggle
6 small
7 overlapping (you may need to
16 tottal draw these
separately to be
able to count each
one. see above.)
5.
6.
Honors Lesson 6Lesson 6
1.
2.
3. triangles, squares,
trapezoids, penntagons
will vary4.
5.
answers
P X X
P X X
= + ( )= +
6 5 6
6 3
.
PP X
P X
P
P
=
=
= ( )=
9
9
9 8
72
6.
$
Lesson 61.
2.
3. triangles, squares,
trapezoids, penntagons
will vary4.
5.
answers
P X X
P X X
= + ( )= +
6 5 6
6 3
.
PP X
P X
P
P
=
=
= ( )=
9
9
9 8
72
6.
$
Honors Lesson 7Lesson 71. extend
co
all segments
aD XY Bc
aB rs Dc
rrresponding
Yes
angles
are congruent
extend 2. ; DDF and Bc
2 line segements are
cut by tran
these
ssversal aB
aDF and
aBe are bot
corresponding s∠ '
hh 90
Dc to include point G
m a = 100
º
º
s
3. extend
∠
iin
º.
ce
m eD
aB and Dc are parallel,
m GDa is 100∠
∠ FF is 80 , since it is
supplementary to GDa.
m D
º
∠
∠ eeF = 90 - definition
of perpendicular
º
4. caB = 90ºº
º
of bisector
aDB = 9
given
BaD definition
( )= −45
00 - definition
of perpendicular
aBD = 45 - fr
º
º oom information given
DBe = 135 - supplementaryº angles
all other corners work out
the same way..
a B
e
G D F c
Honors Lesson 7 - Honors Lesson 8
soLUtions GeoMetrY210
Lesson 71. extend
co
all segments
aD XY Bc
aB rs Dc
rrresponding
Yes
angles
are congruent
extend 2. ; DDF and Bc
2 line segements are
cut by tran
these
ssversal aB
aDF and
aBe are bot
corresponding s∠ '
hh 90
Dc to include point G
m a = 100
º
º
s
3. extend
∠
iin
º.
ce
m eD
aB and Dc are parallel,
m GDa is 100∠
∠ FF is 80 , since it is
supplementary to GDa.
m D
º
∠
∠ eeF = 90 - definition
of perpendicular
º
4. caB = 90ºº
º
of bisector
aDB = 9
given
BaD definition
( )= −45
00 - definition
of perpendicular
aBD = 45 - fr
º
º oom information given
DBe = 135 - supplementaryº angles
all other corners work out
the same way..
e a B
D c
Honors Lesson 8Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
p
o
F
r
lB
n
f
q
e
kh
a d
j
b D
a
c
m i
gc
e
Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
Honors Lesson 8 - Honors Lesson 9
soLUtionsGeoMetrY 211
Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
c
G
e F
a
d D
a
B
b c g
e f
i
h m j
l
k
Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
F
G
D H a
B c
Lesson 81. Look at the drawing below to
see how tthe angles are labeled
for easy reference.
a and d are 25
definition of bisector
p and o are 20
º
ºº
º
definition of bisector
i and j are 45
definitioon of bisector
now look at triangle aeB. its anggles
must add up to 180 We know the
measure
º .
oof a and that of aBc. add
these together, and ssubtract the
result from the total 180 that aº rre
in a triangle:
180 25 90 180 115 65
65
− +( ) = − =
=
º
ºi
Ussing similar reasoning, and looking
at trianglees aec, BFc, aBF, DBc and
aDc, we can find the following:
115
we
m
r f
b g
now
=
= =
= =
º
º º
º . º
95 85
110 70
kknow two angles from each
of the smaller trianggles. armed with
this knowledge, and the fact tthat
there are 180 in a triangle, we can
find
º
the remaining angles:
45c e
q n
k
= =
= =
=
º º
º º
70
65 45
70ºº ºh
You
= 65
can also use what you
know about vertiical angles
and complementary angles
to find somee of the angles.
b, d, j and k are all 902. º defiinition
c a b
of
perpendicular
in a = − +( )180 180º º
ttriangle
180
in
c
i K m
= − +( ) =
= − +( )º º
º º
60 90 30
180 180 aa
triangle
i=180
eGc
i
º º
º
− +( ) =
+ =
90 30 60
90i i angle
ss 90
because
of the
definition of
perpendicul
º
aar.
60 + i = 90
= 30
in a
tr
º
º
º
i
h i j= − +( ) ϒ180 180
iiangle
Bec is
h
h
f h angle
= − +( )=
+ =
180 30 90
60
180
º
º
º
1180
aGe is
90 bec
º
º
º
º
º
f
f
c g angle
+ =
=
+ =
60 180
120
90
aause
of the
definition of
perpendicular.
30 90+ =g ºº
ºg = 60
3. Use the same process for this one.
remeember that you can also use
what you know aboutt vertical angles
or complementary and supplemeentary
angles as a shortcut.
=
=
=
=
22 5
45
67 5
9
. º
º
. º
00
112 5
135
º
. º
º
=
=
Honors Lesson 9Lesson 91. large rectangle:
15' 6" = 15.5 ft
15.5 xx 13 = 201.5 ft2
rectangle:
3 x 5 = 15 ft2
small
ll earg
( ) (
trapezoid:
9( ) + = ( )10 42
9 142
))
( )
= ( ) ( ) =
( ) +
9 7 63 2
4 82
small trapezoid:
2
ft
== ( ) = ( ) ( ) =2 122
2 6 12( ) ft2
total:
201.5 + 15 ++ 63 + 12 = 291.5 ft2
it is necessary sometim2. ees to add
lines to the drawing to make it
clearrer. in figure 1a, dotted lines
have been addedd to show how one
end of the figure has been brroken
up. since we know that the long
measuremeent is 6.40 in and the
space between the dottedd lined is
80 in, we can see that the heights.
of the trapezoids add up to 5.60 in.
since we have been told that the top
and bottom are thee same, each
trapezoid must have a height of
2..80 in.
area of each trapezoid:
2 8 1 27 802
. ( . .( ) + = ( ) =) . ( . )
.
2 8 2 072
2 898 in2
theresince are four trapezoids in all,
we multiply by 4:
22.898 x 4 = 11.592 in2
center portire tanc gular oon:
80 in x 15 in = 12 in2.
:
. .
total
12 11 592 23 5+ = 992 in2
or ab see figure 23.
4.
area a b
area
= ( )( ) ( )== ( )( ) ( )= ( )( )
2 2a b
area na nb
or 4ab see figure 3
5. or n2 see figure 4ab
area n ab
( )= = ( )( ) ( ) =6. 2 52 4 5 225 20
500
( ) ( ) =
ft2
triangle: a = 12
7. first xy
osec nnd x y xy triangle: a = 12
times 12
s
2 2 2
4 2
( )( ) =
= , oo new area is
four times as great.
squa8. first rre: x
square: x2
( )( ) =
( )( ) =
x x
ond x x
2
2 4sec
Honors Lesson 9 - Honors Lesson 9
soLUtions GeoMetrY212
Lesson 91. large rectangle:
15' 6" = 15.5 ft
15.5 xx 13 = 201.5 ft2
rectangle:
3 x 5 = 15 ft2
small
ll earg
( ) (
trapezoid:
9( ) + = ( )10 42
9 142
))
( )
= ( ) ( ) =
( ) +
9 7 63 2
4 82
small trapezoid:
2
ft
== ( ) = ( ) ( ) =2 122
2 6 12( ) ft2
total:
201.5 + 15 ++ 63 + 12 = 291.5 ft2
it is necessary sometim2. ees to add
lines to the drawing to make it
clearrer. in figure 1a, dotted lines
have been addedd to show how one
end of the figure has been brroken
up. since we know that the long
measuremeent is 6.40 in and the
space between the dottedd lined is
80 in, we can see that the heights.
of the trapezoids add up to 5.60 in.
since we have been told that the top
and bottom are thee same, each
trapezoid must have a height of
2..80 in.
area of each trapezoid:
2 8 1 27 802
. ( . .( ) + = ( ) =) . ( . )
.
2 8 2 072
2 898 in2
theresince are four trapezoids in all,
we multiply by 4:
22.898 x 4 = 11.592 in2
center portire tanc gular oon:
80 in x 15 in = 12 in2.
:
. .
total
12 11 592 23 5+ = 992 in2
or ab see figure 23.
4.
area a b
area
= ( )( ) ( )== ( )( ) ( )= ( )( )
2 2a b
area na nb
or 4ab see figure 3
5. or n2 see figure 4ab
area n ab
( )= = ( )( ) ( ) =6. 2 52 4 5 225 20
500
( ) ( ) =
ft2
triangle: a = 12
7. first xy
osec nnd x y xy triangle: a = 12
times 12
s
2 2 2
4 2
( )( ) =
= , oo new area is
four times as great.
squa8. first rre: x
square: x2
( )( ) =
( )( ) =
x x
ond x x
2
2 4sec
figure 1a
figure 1b (shows a different way
of finnding the area
of large rectangle 1
)
area 55 x 6.4 = 96 in2
trapezoid long base
2 x .
one
15− 88 in2
base
2 x 1.27
in2
( ) =
− ( ) =
13 4
15
12 46
.
.
short
heeight
area
6.4 - .8
in2
of one trapezoi
( )
=
÷
2 2 8.
dd
= 36.204 in2
trapezoids 2 x 36.204 =
72.4
Both
008 in2
of figure 96 72.408 =
23.592 in2
area −
trapezoid
trapezoid
6.40"
2.80"
} }}figure 1a
figure 1b (shows a different way
of finnding the area
of large rectangle 1
)
area 55 x 6.4 = 96 in2
trapezoid long base
2 x .
one
15− 88 in2
base
2 x 1.27
in2
( ) =
− ( ) =
13 4
15
12 46
.
.
short
heeight
area
6.4 - .8
in2
of one trapezoi
( )
=
÷
2 2 8.
dd
= 36.204 in2
trapezoids 2 x 36.204 =
72.4
Both
008 in2
of figure 96 72.408 =
23.592 in2
area −
figure 1a
figure 1b (shows a different way
of finnding the area
of large rectangle 1
)
area 55 x 6.4 = 96 in2
trapezoid long base
2 x .
one
15− 88 in2
base
2 x 1.27
in2
( ) =
− ( ) =
13 4
15
12 46
.
.
short
heeight
area
6.4 - .8
in2
of one trapezoi
( )
=
÷
2 2 8.
dd
= 36.204 in2
trapezoids 2 x 36.204 =
72.4
Both
008 in2
of figure 96 72.408 =
23.592 in2
area −
figure 3
2a
2b
figure 2
a
b
Honors Lesson 9 - Honors Lesson 12
soLUtionsGeoMetrY 213
2x x
figure 5
2y y
figure 4
na
nb
x2
figure 6
x
Honors Lesson 10Lesson101- 4.
5. your answer should be close
to 0.661803.
illustration above.
the ratio shoul
6. see
dd be close to what
you got in #5.
7 - 8.
G F e
a D B c
Lesson101- 4.
5. your answer should be close
to 0.661803.
illustration above.
the ratio shoul
6. see
dd be close to what
you got in #5.
7 - 8.
Honors Lesson 11Lesson111.
green, blue,
buttons
green red
zipper z
, ,
iipper buttons
chris x
Douglas x yes x x
ashley x x x y
yes x x
ees
naomi x x yes x
refresh place
2.
planning birthday
ga
−
mmes ments for party
x x yes
yes
guest
sam x
Jason x x x
shanee x x x
troy x x x
boat airplane
Janelle
yes
yes
train car
3.
yyes x x
yes
x
x
chic
x
Walter x x x
Julie yes x x
Jared x yes x
4.
kken tossedhot dog soup salad
yes x x
pizza
Molly x
tinaa x x x
Logan x yes x
sam x x x
answers
yes
x
yes
will vary5. ..
Honors Lesson 12Lesson12
1.
2.
3.
60º
since the sections are all equual,
the center angles are all the
same. 360º ÷88 45= º
º
4 - 8.
9. in #3, you divided 360 by 8
to findd that each small triangle
has a central angle of 45 since
a hexagon has six sides, you wa
º.
nnt
to construct six triangles inside the
circlee. 360
#1, you learned how to
constru
º º÷6 60=
in
cct an equilateral triangle with
each angle equaal to 60 after drawing
a circle and one diam
º.
eeter, use the same
procedure to construct equillateral
triangles inside your circle, using a rradius
of the circle as your starting point eacch
time. after you have constructed four
trianggles,connect their points, and you
will have ann inscribed regular hexagon.
10 -12.
Honors Lesson 12 - Honors Lesson 13
soLUtions GeoMetrY214
Lesson121.
2.
3.
60º
since the sections are all equual,
the center angles are all the
same. 360º ÷88 45= º
º
4 - 8.
9. in #3, you divided 360 by 8
to findd that each small triangle
has a central angle of 45 since
a hexagon has six sides, you wa
º.
nnt
to construct six triangles inside the
circlee. 360
#1, you learned how to
constru
º º÷6 60=
in
cct an equilateral triangle with
each angle equaal to 60 after drawing
a circle and one diam
º.
eeter, use the same
procedure to construct equillateral
triangles inside your circle, using a rradius
of the circle as your starting point eacch
time. after you have constructed four
trianggles,connect their points, and you
will have ann inscribed regular hexagon.
10 -12.
Lesson121.
2.
3.
60º
since the sections are all equual,
the center angles are all the
same. 360º ÷88 45= º
º
4 - 8.
9. in #3, you divided 360 by 8
to findd that each small triangle
has a central angle of 45 since
a hexagon has six sides, you wa
º.
nnt
to construct six triangles inside the
circlee. 360
#1, you learned how to
constru
º º÷6 60=
in
cct an equilateral triangle with
each angle equaal to 60 after drawing
a circle and one diam
º.
eeter, use the same
procedure to construct equillateral
triangles inside your circle, using a rradius
of the circle as your starting point eacch
time. after you have constructed four
trianggles,connect their points, and you
will have ann inscribed regular hexagon.
10 -12.
Lesson121.
2.
3.
60º
since the sections are all equual,
the center angles are all the
same. 360º ÷88 45= º
º
4 - 8.
9. in #3, you divided 360 by 8
to findd that each small triangle
has a central angle of 45 since
a hexagon has six sides, you wa
º.
nnt
to construct six triangles inside the
circlee. 360
#1, you learned how to
constru
º º÷6 60=
in
cct an equilateral triangle with
each angle equaal to 60 after drawing
a circle and one diam
º.
eeter, use the same
procedure to construct equillateral
triangles inside your circle, using a rradius
of the circle as your starting point eacch
time. after you have constructed four
trianggles,connect their points, and you
will have ann inscribed regular hexagon.
10 -12.
Honors Lesson 13Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Honors Lesson 13 - Honors Lesson 14
soLUtionsGeoMetrY 215
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Lesson131.
2.
see illustration.
3 x 9 = 27 units2
33. 12
2 2 2 2
12
3 1 1 5 2
1 6 3
x
x
x
( ) =
( ) =
( ) =
units
units
12
.
uunits
12
units
units
27
2
3 2 3 2
2 1 5 3 3 9 5 2
x( ) =
+ + + =4. . .
- 9.5 = 17.5 units
see illustration for 5 &
2
5. 6.
see illustration for 5 & 6.
units
6.
7. 4 8 32x = 22
1 2 2 2
1 3 1 5 2
2 3 3
8. x
x
x
=
( ) =
( ) =
units
12
units
12
uni
.
tts
12
units
12
units
12
2
2 2 2 2
1 7 3 5 2
1 2 1
x
x
x
( ) =
( ) =
( ) =
.
unit
units
32 - 13 = 19
2
2 1 5 3 2 3 5 1 13 29. + + + + + =. .
uunits
x
2
10 5 50
10. see illustration.
= uunits
12
units
1 x 5 = 5 units
12
2
5 2 5 2
2
1 5
x
x
( ) =
( ) = 22 5 2
4 1 2 2
2
6 1
. units
12
units
1 x 1 = 1 unit
12
x
x
( ) =
(( ) =
( ) =
+ + + + + + =
3 2
3 4 6 2
5 5 2 5 2 1 3 6 24
units
12
unitsx
. .55 2
2
units
50 - 24.5 = 25.5 units
see il11. llustration.
x 8 = 32 units
12
4 2
1 1x( ) ==
( ) =
( )
.
.
5 2
3 1 1 5 2
2
1 2
units
12
units
1x1=1 unit
12
x
x ==
( ) =
( ) =
1 2
1 2 1 2
2
12
5 1
unit
12
unit
1 x 1 = 1 unit
x
x 22 5 2
2 1 1 2
. units
12
unit
.5 + 1.5 + 1 + 1 +1 +
x( ) =
1 + 2.5 + 1
= 9.5 units
32 - 9.5 = 22.5 units
2
2
Honors Lesson 14Lesson141.
2.
12
3 4 12
12 6 2x
a s s a s b
( ) = ( ) =
= −( ) −
units
(( ) −( )
= −( ) −( ) −( )
= ( ) ( )( )=
=
s c
a
a
a
a
6 6 3 6 4 6 5
6 3 2 1
36
6 unnits2
16 16 7 16 10 16 15
16 9 6 1
yes
a
a
3. = −( ) −( ) −( )
= ( ) ( ) ( ))=
=
= −( ) −( ) −(
a
a
a
864
29 39 2
52 52 36 52 28 52 40
. units
4. ))
= ( ) ( )( )=
=
=
a
a
a
V
52 16 24 12
239 616
489 51 2
,
. units
5. πrr h
V
V
V
V
2
3 14 22
10
3 14 4 10
125 6
= ( ) ( )= ( ) ( )
=
=
.
.
. in3
6. πrr h
V
V
V
2
3 14 12
10
3 14 1 10
31 4
= ( ) ( )= ( ) ( )
=
.
.
. in3
it is 114
the first one
7. V r h
V
V
=
= ( ) ( )= ( )
π 2
3 14 22
5
3 14 4
.
. 55
62 8
2
( )
=
=
V
it
V r h
. in3
is half the first one.
8. π
VV
V
V
= ( ) ( )= ( ) ( )
=
3 14 42
10
3 14 16 10
502 4
.
.
. in3
it is foour times the first one.
9. V r h
V
=
= ( ) (
π 2
3 14 22
20. ))= ( ) ( )
=
V
V
3 14 4 20
251 2
.
. cu in3
it is two times thee first one
the height is doubled, the 10. When
vvolume is doubled. When the
height is halved, tthe volume is
halved. When the radius is doubleed,
the volume increases by a factor of 4.
When tthe radius is halved, the volume
decreases by aa factor of 4.
the student may use his own wordss
to express this.
answers will vary.
take
11.
12. the formula, and multiply both
sides by 2:
V r= π 22
rearrange the factors:
V r2h
2V
h
V r h
now
2 2 2=
=
=
π
π
πrr h
take
22
the formula, and multiply
both sides bby 4:
V
4V 4
the 4 on the right
=
=
π
π
r h
r h
write
2
2
re sside as 22:
V 22
the factors:
V
4 2
4
=
=
π
π
r h
arrangere
222 2
4 22
r h
r hV
there is more than one way to s
= ( )π
eet
this up. as long as you show the
same resultts as by experimentation,
the answer is correct..
Honors Lesson 14 - Honors Lesson 14
soLUtions GeoMetrY216
Lesson141.
2.
12
3 4 12
12 6 2x
a s s a s b
( ) = ( ) =
= −( ) −
units
(( ) −( )
= −( ) −( ) −( )
= ( ) ( )( )=
=
s c
a
a
a
a
6 6 3 6 4 6 5
6 3 2 1
36
6 unnits2
16 16 7 16 10 16 15
16 9 6 1
yes
a
a
3. = −( ) −( ) −( )
= ( ) ( ) ( ))=
=
= −( ) −( ) −(
a
a
a
864
29 39 2
52 52 36 52 28 52 40
. units
4. ))
= ( ) ( )( )=
=
=
a
a
a
V
52 16 24 12
239 616
489 51 2
,
. units
5. πrr h
V
V
V
V
2
3 14 22
10
3 14 4 10
125 6
= ( ) ( )= ( ) ( )
=
=
.
.
. in3
6. πrr h
V
V
V
2
3 14 12
10
3 14 1 10
31 4
= ( ) ( )= ( ) ( )
=
.
.
. in3
it is 114
the first one
7. V r h
V
V
=
= ( ) ( )= ( )
π 2
3 14 22
5
3 14 4
.
. 55
62 8
2
( )
=
=
V
it
V r h
. in3
is half the first one.
8. π
VV
V
V
= ( ) ( )= ( ) ( )
=
3 14 42
10
3 14 16 10
502 4
.
.
. in3
it is foour times the first one.
9. V r h
V
=
= ( ) (
π 2
3 14 22
20. ))= ( ) ( )
=
V
V
3 14 4 20
251 2
.
. cu in3
it is two times thee first one
the height is doubled, the 10. When
vvolume is doubled. When the
height is halved, tthe volume is
halved. When the radius is doubleed,
the volume increases by a factor of 4.
When tthe radius is halved, the volume
decreases by aa factor of 4.
the student may use his own wordss
to express this.
answers will vary.
take
11.
12. the formula, and multiply both
sides by 2:
V r= π 22
rearrange the factors:
V r2h
2V
h
V r h
now
2 2 2=
=
=
π
π
πrr h
take
22
the formula, and multiply
both sides bby 4:
V
4V 4
the 4 on the right
=
=
π
π
r h
r h
write
2
2
re sside as 22:
V 22
the factors:
V
4 2
4
=
=
π
π
r h
arrangere
222 2
4 22
r h
r hV
there is more than one way to s
= ( )π
eet
this up. as long as you show the
same resultts as by experimentation,
the answer is correct..
Lesson141.
2.
12
3 4 12
12 6 2x
a s s a s b
( ) = ( ) =
= −( ) −
units
(( ) −( )
= −( ) −( ) −( )
= ( ) ( )( )=
=
s c
a
a
a
a
6 6 3 6 4 6 5
6 3 2 1
36
6 unnits2
16 16 7 16 10 16 15
16 9 6 1
yes
a
a
3. = −( ) −( ) −( )
= ( ) ( ) ( ))=
=
= −( ) −( ) −(
a
a
a
864
29 39 2
52 52 36 52 28 52 40
. units
4. ))
= ( ) ( )( )=
=
=
a
a
a
V
52 16 24 12
239 616
489 51 2
,
. units
5. πrr h
V
V
V
V
2
3 14 22
10
3 14 4 10
125 6
= ( ) ( )= ( ) ( )
=
=
.
.
. in3
6. πrr h
V
V
V
2
3 14 12
10
3 14 1 10
31 4
= ( ) ( )= ( ) ( )
=
.
.
. in3
it is 114
the first one
7. V r h
V
V
=
= ( ) ( )= ( )
π 2
3 14 22
5
3 14 4
.
. 55
62 8
2
( )
=
=
V
it
V r h
. in3
is half the first one.
8. π
VV
V
V
= ( ) ( )= ( ) ( )
=
3 14 42
10
3 14 16 10
502 4
.
.
. in3
it is foour times the first one.
9. V r h
V
=
= ( ) (
π 2
3 14 22
20. ))= ( ) ( )
=
V
V
3 14 4 20
251 2
.
. cu in3
it is two times thee first one
the height is doubled, the 10. When
vvolume is doubled. When the
height is halved, tthe volume is
halved. When the radius is doubleed,
the volume increases by a factor of 4.
When tthe radius is halved, the volume
decreases by aa factor of 4.
the student may use his own wordss
to express this.
answers will vary.
take
11.
12. the formula, and multiply both
sides by 2:
V r= π 22
rearrange the factors:
V r2h
2V
h
V r h
now
2 2 2=
=
=
π
π
πrr h
take
22
the formula, and multiply
both sides bby 4:
V
4V 4
the 4 on the right
=
=
π
π
r h
r h
write
2
2
re sside as 22:
V 22
the factors:
V
4 2
4
=
=
π
π
r h
arrangere
222 2
4 22
r h
r hV
there is more than one way to s
= ( )π
eet
this up. as long as you show the
same resultts as by experimentation,
the answer is correct..
Lesson141.
2.
12
3 4 12
12 6 2x
a s s a s b
( ) = ( ) =
= −( ) −
units
(( ) −( )
= −( ) −( ) −( )
= ( ) ( )( )=
=
s c
a
a
a
a
6 6 3 6 4 6 5
6 3 2 1
36
6 unnits2
16 16 7 16 10 16 15
16 9 6 1
yes
a
a
3. = −( ) −( ) −( )
= ( ) ( ) ( ))=
=
= −( ) −( ) −(
a
a
a
864
29 39 2
52 52 36 52 28 52 40
. units
4. ))
= ( ) ( )( )=
=
=
a
a
a
V
52 16 24 12
239 616
489 51 2
,
. units
5. πrr h
V
V
V
V
2
3 14 22
10
3 14 4 10
125 6
= ( ) ( )= ( ) ( )
=
=
.
.
. in3
6. πrr h
V
V
V
2
3 14 12
10
3 14 1 10
31 4
= ( ) ( )= ( ) ( )
=
.
.
. in3
it is 114
the first one
7. V r h
V
V
=
= ( ) ( )= ( )
π 2
3 14 22
5
3 14 4
.
. 55
62 8
2
( )
=
=
V
it
V r h
. in3
is half the first one.
8. π
VV
V
V
= ( ) ( )= ( ) ( )
=
3 14 42
10
3 14 16 10
502 4
.
.
. in3
it is foour times the first one.
9. V r h
V
=
= ( ) (
π 2
3 14 22
20. ))= ( ) ( )
=
V
V
3 14 4 20
251 2
.
. cu in3
it is two times thee first one
the height is doubled, the 10. When
vvolume is doubled. When the
height is halved, tthe volume is
halved. When the radius is doubleed,
the volume increases by a factor of 4.
When tthe radius is halved, the volume
decreases by aa factor of 4.
the student may use his own wordss
to express this.
answers will vary.
take
11.
12. the formula, and multiply both
sides by 2:
V r= π 22
rearrange the factors:
V r2h
2V
h
V r h
now
2 2 2=
=
=
π
π
πrr h
take
22
the formula, and multiply
both sides bby 4:
V
4V 4
the 4 on the right
=
=
π
π
r h
r h
write
2
2
re sside as 22:
V 22
the factors:
V
4 2
4
=
=
π
π
r h
arrangere
222 2
4 22
r h
r hV
there is more than one way to s
= ( )π
eet
this up. as long as you show the
same resultts as by experimentation,
the answer is correct..
Honors Lesson 15 - Honors Lesson 15
soLUtionsGeoMetrY 217
Honors Lesson 15Lesson151.
2.
3 3 27 x 3 x ft
12 x 12 x 12 1,728
=
= iin3
x 4 x 2 64 in3
x .3 19.2 lb
in3 1
3.
4.
8
64
64
=
=
÷ ,,728 .037 ft3
x 1200 = 44.4 lbs
You could p
=
.037
rrobably lift it,
but it would be much heavier
thaan expected.
First find what the volume would5.
bbe if it were solid:
V r2=
= ( ) ( )
=
π h
V
V
3 14 52
12
9 4
. .
. 22 in3
find the volume inside
the pipe:
V r2
now
= π hh
V
V
= ( ) ( )
=
3 14 252
12
2 355
. .
. in3
then find the diffeerence:
9.42 2.355 = 7.065 in3−
× =6. 7 065 26 1 836. . . 99
43
3
43
3 14 253
07
lb
in3
7. V r
V
V rounded
=
= ( ) ( )
= (
π
. .
. ))×
=
. . .07 3 02≈
÷
pounds for
one bearing
25 .02 1,250 bearings
we rounded some
numbers, the
Because
aactual number
of bearings in the box may be
sliightly different. Keep in mind
that the startinng weight was
rounded to a whole number.
our annswer is close enough to
be helpful in a real llife situation,
where someone to know
ap
wants
pproximately how many bearings
are available witthout counting.
the side view is a trapezoid,8.
aand the volume of the water
is the area of the ttrapezoid
times the width of the pool:
a 3+102
= 40(( )
= ( )
=
= ( )
=
a
a
V
V
6 5 40
260
260 20
5 200
.
,
ft2
ft3
Volu9. mme of the sphere:
V 43
units
V 4.19
Vo
= ( ) ( )
=
3 14 13 3.
llume of the cube:
V 2 x 2 x 2 8 units
8 4.19 =
= =
−
3
33.81 units
Volume of the cylinder:
V 3.14 1
3
10.
= ( )) ( )
=
22
6 28 3V . units
Volume of the sphere from #9::
4.19 units
6.28 4.19 2.09 units
note: You may
3
3− =
use the
fractional value of if
it seems more
π
convenient.
Lesson151.
2.
3 3 27 x 3 x ft
12 x 12 x 12 1,728
=
= iin3
x 4 x 2 64 in3
x .3 19.2 lb
in3 1
3.
4.
8
64
64
=
=
÷ ,,728 .037 ft3
x 1200 = 44.4 lbs
You could p
=
.037
rrobably lift it,
but it would be much heavier
thaan expected.
First find what the volume would5.
bbe if it were solid:
V r2=
= ( ) ( )
=
π h
V
V
3 14 52
12
9 4
. .
. 22 in3
find the volume inside
the pipe:
V r2
now
= π hh
V
V
= ( ) ( )
=
3 14 252
12
2 355
. .
. in3
then find the diffeerence:
9.42 2.355 = 7.065 in3−
× =6. 7 065 26 1 836. . . 99
43
3
43
3 14 253
07
lb
in3
7. V r
V
V rounded
=
= ( ) ( )
= (
π
. .
. ))×
=
. . .07 3 02≈
÷
pounds for
one bearing
25 .02 1,250 bearings
we rounded some
numbers, the
Because
aactual number
of bearings in the box may be
sliightly different. Keep in mind
that the startinng weight was
rounded to a whole number.
our annswer is close enough to
be helpful in a real llife situation,
where someone to know
ap
wants
pproximately how many bearings
are available witthout counting.
the side view is a trapezoid,8.
aand the volume of the water
is the area of the ttrapezoid
times the width of the pool:
a 3+102
= 40(( )
= ( )
=
= ( )
=
a
a
V
V
6 5 40
260
260 20
5 200
.
,
ft2
ft3
Volu9. mme of the sphere:
V 43
units
V 4.19
Vo
= ( ) ( )
=
3 14 13 3.
llume of the cube:
V 2 x 2 x 2 8 units
8 4.19 =
= =
−
3
33.81 units
Volume of the cylinder:
V 3.14 1
3
10.
= ( )) ( )
=
22
6 28 3V . units
Volume of the sphere from #9::
4.19 units
6.28 4.19 2.09 units
note: You may
3
3− =
use the
fractional value of if
it seems more
π
convenient.
Honors Lesson 15 - Honors Lesson 18
soLUtions GeoMetrY218
Lesson151.
2.
3 3 27 x 3 x ft
12 x 12 x 12 1,728
=
= iin3
x 4 x 2 64 in3
x .3 19.2 lb
in3 1
3.
4.
8
64
64
=
=
÷ ,,728 .037 ft3
x 1200 = 44.4 lbs
You could p
=
.037
rrobably lift it,
but it would be much heavier
thaan expected.
First find what the volume would5.
bbe if it were solid:
V r2=
= ( ) ( )
=
π h
V
V
3 14 52
12
9 4
. .
. 22 in3
find the volume inside
the pipe:
V r2
now
= π hh
V
V
= ( ) ( )
=
3 14 252
12
2 355
. .
. in3
then find the diffeerence:
9.42 2.355 = 7.065 in3−
× =6. 7 065 26 1 836. . . 99
43
3
43
3 14 253
07
lb
in3
7. V r
V
V rounded
=
= ( ) ( )
= (
π
. .
. ))×
=
. . .07 3 02≈
÷
pounds for
one bearing
25 .02 1,250 bearings
we rounded some
numbers, the
Because
aactual number
of bearings in the box may be
sliightly different. Keep in mind
that the startinng weight was
rounded to a whole number.
our annswer is close enough to
be helpful in a real llife situation,
where someone to know
ap
wants
pproximately how many bearings
are available witthout counting.
the side view is a trapezoid,8.
aand the volume of the water
is the area of the ttrapezoid
times the width of the pool:
a 3+102
= 40(( )
= ( )
=
= ( )
=
a
a
V
V
6 5 40
260
260 20
5 200
.
,
ft2
ft3
Volu9. mme of the sphere:
V 43
units
V 4.19
Vo
= ( ) ( )
=
3 14 13 3.
llume of the cube:
V 2 x 2 x 2 8 units
8 4.19 =
= =
−
3
33.81 units
Volume of the cylinder:
V 3.14 1
3
10.
= ( )) ( )
=
22
6 28 3V . units
Volume of the sphere from #9::
4.19 units
6.28 4.19 2.09 units
note: You may
3
3− =
use the
fractional value of if
it seems more
π
convenient.
Honors Lesson 16Lesson161.
2.
r r r
a LW LW LH LH WH WH
LW
( ) =
= + + + + +
= +
π π 2
2 2LLH WH
LW LH WH
s s s s s
V
+
= + +( )
+ +( ) = ( ) =
=
2
2
2 2 2 2 2 3 2 6 23.
4. 33 11 3 99
2 3 11 2 3 3 2 11 3
2 33
( ) ( ) =
= ( ) + ( ) + ( )= ( )
ft3
sa x x x
++ ( ) + ( )= + +
=
=
2 9 2 33
66 18 66
150
150 6
ft2
ft2 faces5. ÷ 225 ft2 per face
25 ft
the new bin is 5 x 5 x 5
= 5
..
cube-shaped one holds more.
125 99 = 26
6. the
− fft3 difference.
Honors Lesson 17Lesson171.
2.
V r h
V
V
V
=
= ( ) ( )
=
=
π 2
3 14 22
4
50 24
43
.
. ft3
ππr
V
V rounded
V
3
43
3 14 23
33 49
3 14 3
= ( ) ( )
= ( )
=
.
.
.
ft3
3. (( ) ( )
=
= ( ) ( )
=
26
169 56 3
43
3 14 33
113 04
V
V
V
.
.
.
units
4.
uunits rounded
units
3
3 14 12
2
6 28 3
( )
= ( ) ( )
=
5.
6.
V
V
.
.
VV
V
= ( ) ( )
= ( )
43
3 14 13
4 19 3
33 4950
.
.
.
units rounded
7...
. ..
.
.
..
2467 113 04
169 5667
4 196 28
67
23
2
≈ ≈
≈
π
8.
9. a r= 22 2
2 3 14 32
2 3 14 3 6
56 52 113 0
+
= ( ) ( ) + ( ) ( ) ( )
= +
πrh
a
a
. .
. . 44 169 56
4 3 14 32
113 04 2
=
= ( ) ( )
=
.
.
.
units2
units
10. a
a
111.
12.
113 04169 56
23
.
.≈
the surface area and volumme of a
sphere appear to be 23
of the
surface aarea and volume of a
cylinder with the same dimmensions.
(archimedes proved that this is
the ccase.)
Lesson171.
2.
V r h
V
V
V
=
= ( ) ( )
=
=
π 2
3 14 22
4
50 24
43
.
. ft3
ππr
V
V rounded
V
3
43
3 14 23
33 49
3 14 3
= ( ) ( )
= ( )
=
.
.
.
ft3
3. (( ) ( )
=
= ( ) ( )
=
26
169 56 3
43
3 14 33
113 04
V
V
V
.
.
.
units
4.
uunits rounded
units
3
3 14 12
2
6 28 3
( )
= ( ) ( )
=
5.
6.
V
V
.
.
VV
V
= ( ) ( )
= ( )
43
3 14 13
4 19 3
33 4950
.
.
.
units rounded
7...
. ..
.
.
..
2467 113 04
169 5667
4 196 28
67
23
2
≈ ≈
≈
π
8.
9. a r= 22 2
2 3 14 32
2 3 14 3 6
56 52 113 0
+
= ( ) ( ) + ( ) ( ) ( )
= +
πrh
a
a
. .
. . 44 169 56
4 3 14 32
113 04 2
=
= ( ) ( )
=
.
.
.
units2
units
10. a
a
111.
12.
113 04169 56
23
.
.≈
the surface area and volumme of a
sphere appear to be 23
of the
surface aarea and volume of a
cylinder with the same dimmensions.
(archimedes proved that this is
the ccase.)
Lesson171.
2.
V r h
V
V
V
=
= ( ) ( )
=
=
π 2
3 14 22
4
50 24
43
.
. ft3
ππr
V
V rounded
V
3
43
3 14 23
33 49
3 14 3
= ( ) ( )
= ( )
=
.
.
.
ft3
3. (( ) ( )
=
= ( ) ( )
=
26
169 56 3
43
3 14 33
113 04
V
V
V
.
.
.
units
4.
uunits rounded
units
3
3 14 12
2
6 28 3
( )
= ( ) ( )
=
5.
6.
V
V
.
.
VV
V
= ( ) ( )
= ( )
43
3 14 13
4 19 3
33 4950
.
.
.
units rounded
7...
. ..
.
.
..
2467 113 04
169 5667
4 196 28
67
23
2
≈ ≈
≈
π
8.
9. a r= 22 2
2 3 14 32
2 3 14 3 6
56 52 113 0
+
= ( ) ( ) + ( ) ( ) ( )
= +
πrh
a
a
. .
. . 44 169 56
4 3 14 32
113 04 2
=
= ( ) ( )
=
.
.
.
units2
units
10. a
a
111.
12.
113 04169 56
23
.
.≈
the surface area and volumme of a
sphere appear to be 23
of the
surface aarea and volume of a
cylinder with the same dimmensions.
(archimedes proved that this is
the ccase.)
Honors Lesson 18Lesson181.
2.
4 003
90
,
º;
mi
a tangent to a circle is
perpendicular to the diameter
3. L2 4 0002 4+ =, ,00032
2 4 0032 4 0002
2 16 024 009 16 000 000
2
L
L
L
= −
= −
, ,
, , , ,
==
=
+
24 009
24 009 155
29 035 5 280 5
2
,
,
, ,
L
L
≈
÷ ≈
mi
mi4.
5. 44 0002 4 0052
2 16 000 000 16 040 025
2 16 040
, ,
, , , ,
,
=
+ =
=
L
L ,, , ,
,
,
025 16 000 000
2 40 025
40 025 200
555 5
−
=
=
L
L ≈
÷
mi
6. ,, .
, , .
, , , ,
280 1
2 4 0002 4 000 12
2 16 000 000 16 000
≈
L
L
+ =
+ = 8800 01
2 16 000 800 01 16 000 000
2 800 01
80
.
, , . , ,
.
L
L
L
= −
=
= 00 01 28 3
1502 4 0002 4 0002
2 8 000
. .
, ,
,
≈ mi
7.
8.
+ = +( )+
X
X XX
X X
+
+
= + +
16 000 000
22 500 16 000 000
2 8 000 16 0
, ,
, , ,
, ,
9.
000 000
22 500 2 8 000
0 2 8 000 22 500
,
, ,
, ,
= +
= + −
+
X X
X X
or X2 88 000 22 500 0
8 000 22 500
22 500 8 000
, ,
, ,
, ,
X
X
X
X
− =
=
=
10.
÷
≈≈ 2 8. mi
Honors Lesson 18 - Honors Lesson 19
soLUtionsGeoMetrY 219
Lesson181.
2.
4 003
90
,
º;
mi
a tangent to a circle is
perpendicular to the diameter
3. L2 4 0002 4+ =, ,00032
2 4 0032 4 0002
2 16 024 009 16 000 000
2
L
L
L
= −
= −
, ,
, , , ,
==
=
+
24 009
24 009 155
29 035 5 280 5
2
,
,
, ,
L
L
≈
÷ ≈
mi
mi4.
5. 44 0002 4 0052
2 16 000 000 16 040 025
2 16 040
, ,
, , , ,
,
=
+ =
=
L
L ,, , ,
,
,
025 16 000 000
2 40 025
40 025 200
555 5
−
=
=
L
L ≈
÷
mi
6. ,, .
, , .
, , , ,
280 1
2 4 0002 4 000 12
2 16 000 000 16 000
≈
L
L
+ =
+ = 8800 01
2 16 000 800 01 16 000 000
2 800 01
80
.
, , . , ,
.
L
L
L
= −
=
= 00 01 28 3
1502 4 0002 4 0002
2 8 000
. .
, ,
,
≈ mi
7.
8.
+ = +( )+
X
X XX
X X
+
+
= + +
16 000 000
22 500 16 000 000
2 8 000 16 0
, ,
, , ,
, ,
9.
000 000
22 500 2 8 000
0 2 8 000 22 500
,
, ,
, ,
= +
= + −
+
X X
X X
or X2 88 000 22 500 0
8 000 22 500
22 500 8 000
, ,
, ,
, ,
X
X
X
X
− =
=
=
10.
÷
≈≈ 2 8. mi
Honors Lesson 19Lesson191. V
V
=
= ⋅( ) ( )area of base x altitude
V 4 4 8
==
= ( ) + ( ) + ( )= ( ) + (
128
2 4 4 2 4 8 2 4 8
2 16 2 32
in3
2. sa x x x
sa )) + ( )= + +
=
=
2 32
32 64 64
160
sa
sa
V
in2
area of base 3. xx altitude
V= 12
ft3
3 4 10
60
2 12
3 4
x x
Y
sa x
( )
=
= ( ) ( )4. + ( ) +
( ) + ( )= + + +
=
3 10
4 10 5 10
12 30 40 50
132
x
x x
sa
sa ft2
think of the wire as a long,
skinny cyli
5.
nnder.
1 ft3 in3
Volume of wire =
= =12 12 12 1 728x x ,
aarea of base
x length
1,728 3.14 x .12 = ( )x L
1 72, 88 0314
55 031 8
55 031 8 12 4 586
=
=
.
, .
, . ,
L
a L
in L
ft
≈
÷ ≈
6. WW Let L the circumference andW the height of
== tthe cylinder.
so L 3.14 9
in
Diameter = = ( )9
28 26
,
. ≈ LL this is one
dimension of the rectangle annd the
circumference of the cylinder.
625 28.26W=
222.12 in W this is the other
dimension of the
=
rectangle and the
height of the cylinder.
V are= aa of base x height
V 3.14 9 2 = ( )
=
÷2
22 12
3 14 4 5
x
V
.
. .(( )2 22 12
1 406 5
x
V
in3
cylinder will be 4 i
.
, .≈
7. nn high and
in3 in diameter. area of one
circu
4
llar end 3.14 2 in2
of side 3.14 4
= ( ) =
= ( )
212 56.
area xx in2
in2
You al
4 50 24
50 24 12 56 12 56 75 36
=
+ + =
.
. . . .
sso could have used what you
learned in lesson 117 to find the
surface area of the cylinder. Fiirst
find the surface area of the sphere,
and tthen multiply by 32
(see
below for an altern
.
aative solution.)
alternative solution
sa of sp
7.
hhere 4 3.14
in2
or 1.5 50.24
= ( ) ( ) =
( ) =
22
50 24
32
7
.
55 36
2 4 4 2 4 4 2 4 4
32 32 32 96
. in2
8. a x x x
a
= ( ) + ( ) + ( )
= + + = iin2
the cylinder uses less cardboard.
(However, tthere will be odd-shaped,
unuseable pipossibly eeces left over.)
Honors Lesson 19 - Honors Lesson 21
soLUtions GeoMetrY220
Lesson191. V
V
=
= ⋅( ) ( )area of base x altitude
V 4 4 8
==
= ( ) + ( ) + ( )= ( ) + (
128
2 4 4 2 4 8 2 4 8
2 16 2 32
in3
2. sa x x x
sa )) + ( )= + +
=
=
2 32
32 64 64
160
sa
sa
V
in2
area of base 3. xx altitude
V= 12
ft3
3 4 10
60
2 12
3 4
x x
Y
sa x
( )
=
= ( ) ( )4. + ( ) +
( ) + ( )= + + +
=
3 10
4 10 5 10
12 30 40 50
132
x
x x
sa
sa ft2
think of the wire as a long,
skinny cyli
5.
nnder.
1 ft3 in3
Volume of wire =
= =12 12 12 1 728x x ,
aarea of base
x length
1,728 3.14 x .12 = ( )x L
1 72, 88 0314
55 031 8
55 031 8 12 4 586
=
=
.
, .
, . ,
L
a L
in L
ft
≈
÷ ≈
6. WW Let L the circumference andW the height of
== tthe cylinder.
so L 3.14 9
in
Diameter = = ( )9
28 26
,
. ≈ LL this is one
dimension of the rectangle annd the
circumference of the cylinder.
625 28.26W=
222.12 in W this is the other
dimension of the
=
rectangle and the
height of the cylinder.
V are= aa of base x height
V 3.14 9 2 = ( )
=
÷2
22 12
3 14 4 5
x
V
.
. .(( )2 22 12
1 406 5
x
V
in3
cylinder will be 4 i
.
, .≈
7. nn high and
in3 in diameter. area of one
circu
4
llar end 3.14 2 in2
of side 3.14 4
= ( ) =
= ( )
212 56.
area xx in2
in2
You al
4 50 24
50 24 12 56 12 56 75 36
=
+ + =
.
. . . .
sso could have used what you
learned in lesson 117 to find the
surface area of the cylinder. Fiirst
find the surface area of the sphere,
and tthen multiply by 32
(see
below for an altern
.
aative solution.)
alternative solution
sa of sp
7.
hhere 4 3.14
in2
or 1.5 50.24
= ( ) ( ) =
( ) =
22
50 24
32
7
.
55 36
2 4 4 2 4 4 2 4 4
32 32 32 96
. in2
8. a x x x
a
= ( ) + ( ) + ( )
= + + = iin2
the cylinder uses less cardboard.
(However, tthere will be odd-shaped,
unuseable pipossibly eeces left over.)
Honors Lesson 20Lesson 201.
2.
300 150 2÷ = hours
answers may vary
the wind blew him off course.
3.
4.
30 2 15
150 2
÷
÷
=
=
mm
775
80
75
mm
see drawing.
answ
5.
6.
∠ =
∠ =
oWP
oWP
Your
º
º
eers to #5 and #6
may vary slightly depending
how carefully you drew and
measured.
P
75º
north
20
mm
o
W
east
south
75 mm
Honors Lesson 21Lesson 211.
2.
3.
4.
π
π π
y
a x y
y z x
z x y
a
2
2 2
2 2 2
2 2 2
= −
+ =
= −
= ππ
π
π
π
π
x y
a z
a z
a
a
2 2
2
2
102
2
5
−( )= ( )= ( )
=
=
5.
6.
( )
(( )=
=
= ( )
= × =
2
3 14 25
78 5
3 14 42
3 14 16
a
a
a
a
.
.
.
.
x
in2
7.
550 24
50 24 007
50 24 007 7 177
.
. .
. . ,
in2
8. a Lx W
L
=
= ×
÷ ≈ iin
tickets
(rounded to the neare
9. 7 177 2 3 589, ,÷ ≈
sst
whole number)
Honors Lesson 22 - Honors Lesson 25
soLUtionsGeoMetrY 221
Honors Lesson 22Lesson 221.
2.
this bird is red.
is congruent t∠a oo B.
i get 100% on my math test.
this trian
∠
3.
4. ggle has two
congruent sides.
Honors Lesson 23Lesson 231. if i get burned, i touched the
hot stoove. not necessarily true.
if two line segmen2. tts are congruent,
they have equal length.
true.
33. if a bird is red, it is a cardinal.
not necesssarily true.
if the leg squared plus the leg
s
4.
qquared equals the hypotenuse
squared, the trianggle is a right
triangle.
true.
if my plants wil5. tt, i stop
watering them.
not true if i am sensiblle!
Honors Lesson 24Lesson 241. 50º; the measure of an inscribed
anglee is half the measure of the
intercepted arc.
2. 1130 180 50
50
80
º; º º
º;
º;
same reason as #1
1
−
3.
4. 880 50 50
ve
º º º
º; º º º
º;
− +( )− +( )5.
6.
160 360 100 100
80 rrtical angles
180
180 80
7.
8.
85 95
15 85
º; º º
º; º º
−
− +(( )checking results with remote
interior angles:: 80
angle 1 and the 70 angle
n
º º º
º; º
+ =15 95
809.
eext to it put together form an
angle that is thee alternate interior
angle to the 150 angle atº the
top left.
150
alternate in
º º º
º;
− =70 80
7010. tterior angles
180
alt
11.
12.
30 70 80
30
º; º º º
º;
− +( )eernate interior angles
Lesson 241. 50º; the measure of an inscribed
anglee is half the measure of the
intercepted arc.
2. 1130 180 50
50
80
º; º º
º;
º;
same reason as #1
1
−
3.
4. 880 50 50
ve
º º º
º; º º º
º;
− +( )− +( )5.
6.
160 360 100 100
80 rrtical angles
180
180 80
7.
8.
85 95
15 85
º; º º
º; º º
−
− +(( )checking results with remote
interior angles:: 80
angle 1 and the 70 angle
n
º º º
º; º
+ =15 95
809.
eext to it put together form an
angle that is thee alternate interior
angle to the 150 angle atº the
top left.
150
alternate in
º º º
º;
− =70 80
7010. tterior angles
180
alt
11.
12.
30 70 80
30
º; º º º
º;
− +( )eernate interior angles
Honors Lesson 25Lesson 251.
statements asons
aF eF GivenGive
re
≅∠ ≅ ∠1 2 nn
cF cF flexiveceF caF sas
ce cacorresponding
≅≅
≅
re
pparts
of congruent triangles
sss cDe cBa
cDe c
≅
∠ ≅ ∠ BBacorresponding
s
parts
of congruent triangles
2.
ttatements asons
tU rQ Given
tUV rQV Given
UV QV G
re
≅
∠ ≅ ∠
≅ iiven
tUV rQV sas
tV rV cPctrc
st sr
sV sV fl
≅
≅
≅
≅
Given
re eexivetsV rsV ssstsV rsV cPctrc
statements
≅∠ ≅ ∠
3.
reaasons
Fe GH Given
FH Ge Given
eH eH flexiveFeH G
≅
≅
≅≅
re HHe sss
D e
F
a B
2
1 c
Honors Lesson 25 - Honors Lesson 27
soLUtions GeoMetrY222
Lesson 251.
statements asons
aF eF GivenGive
re
≅∠ ≅ ∠1 2 nn
cF cF flexiveceF caF sas
ce cacorresponding
≅≅
≅
re
pparts
of congruent triangles
sss cDe cBa
cDe c
≅
∠ ≅ ∠ BBacorresponding
s
parts
of congruent triangles
2.
ttatements asons
tU rQ Given
tUV rQV Given
UV QV G
re
≅
∠ ≅ ∠
≅ iiven
tUV rQV sas
tV rV cPctrc
st sr
sV sV fl
≅
≅
≅
≅
Given
re eexivetsV rsV ssstsV rsV cPctrc
statements
≅∠ ≅ ∠
3.
reaasons
Fe GH Given
FH Ge Given
eH eH flexiveFeH G
≅
≅
≅≅
re HHe sss
V
U
Q r
t
s
Lesson 251.
statements asons
aF eF GivenGive
re
≅∠ ≅ ∠1 2 nn
cF cF flexiveceF caF sas
ce cacorresponding
≅≅
≅
re
pparts
of congruent triangles
sss cDe cBa
cDe c
≅
∠ ≅ ∠ BBacorresponding
s
parts
of congruent triangles
2.
ttatements asons
tU rQ Given
tUV rQV Given
UV QV G
re
≅
∠ ≅ ∠
≅ iiven
tUV rQV sas
tV rV cPctrc
st sr
sV sV fl
≅
≅
≅
≅
Given
re eexivetsV rsV ssstsV rsV cPctrc
statements
≅∠ ≅ ∠
3.
reaasons
Fe GH Given
FH Ge Given
eH eH flexiveFeH G
≅
≅
≅≅
re HHe sss
H G e
F
e H
Honors Lesson 26Lesson 261.
statements asons
aB ac GivenarB aQc
re
≅∠ ≅ ∠ PPerpendicular
Bar caQ flexive
Bar caQ aas
∠ ≅ ∠
≅
re
or Ha
Definiti
cQ Br cPctrc
statements asons
XB YB
≅
≅
2.
re
oon of bisector of Perpendicu∠ ≅ ∠XBa YBa Definiton llar
or LLBa Ba flexive
XBa YBa sas
Xa Ya cPctrc
≅≅
≅
re
33.
statements asons
eF GF
eX GX
re
≅
≅
From proof above
Deefinition of Bisector
FX FX flexiveeFX GFX ss
≅≅
re ss
eXH GXH
HX
or HLDefinition of Perpendicular∠ ≅ ∠
≅ HHXeHX GHX
eH GH
reflexivesas or LL
cPctrc
≅
≅
Lesson 261.
statements asons
aB ac GivenarB aQc
re
≅∠ ≅ ∠ PPerpendicular
Bar caQ flexive
Bar caQ aas
∠ ≅ ∠
≅
re
or Ha
Definiti
cQ Br cPctrc
statements asons
XB YB
≅
≅
2.
re
oon of bisector of Perpendicu∠ ≅ ∠XBa YBa Definiton llar
or LLBa Ba flexive
XBa YBa sas
Xa Ya cPctrc
≅≅
≅
re
33.
statements asons
eF GF
eX GX
re
≅
≅
From proof above
Deefinition of Bisector
FX FX flexiveeFX GFX ss
≅≅
re ss
eXH GXH
HX
or HLDefinition of Perpendicular∠ ≅ ∠
≅ HHXeHX GHX
eH GH
reflexivesas or LL
cPctrc
≅
≅
Honors Lesson 27Lesson 261.
statements asons
Xc Yc radius
re
≅ of a cirrclea tangent of a circle
is perpendic∠ ≅ ∠PYc PXc uular to the
radius at that point.
Pc Pc flexiv≅ re eePYc PXc HL
PX PY cPctrc
statements asons
De a
≅
≅
⊥
2.
re
BB
ac Bc radius
Fc Fc flexiveFc
Given
of a circle≅
≅ re aa FcB HL
ace Bce cPctrc
ae Be
≅∠ ≅ ∠
≅
Property of
centraal angle
Given
radi
3.
statements asons
oP LM
oc Lc
re
≅
≅ uus of a circle
of a circlePc Mc radiuscPo cML
≅≅ ssss
oX LYocX LcY
Xc Yc
≅≅
≅
Definition of BisectorHL
ccPctrc
Honors Lesson 27 - Honors Lesson 29
soLUtionsGeoMetrY 223
P c
Y
X Lesson 26
1.
statements asons
Xc Yc radius
re
≅ of a cirrclea tangent of a circle
is perpendic∠ ≅ ∠PYc PXc uular to the
radius at that point.
Pc Pc flexiv≅ re eePYc PXc HL
PX PY cPctrc
statements asons
De a
≅
≅
⊥
2.
re
BB
ac Bc radius
Fc Fc flexiveFc
Given
of a circle≅
≅ re aa FcB HL
ace Bce cPctrc
ae Be
≅∠ ≅ ∠
≅
Property of
centraal angle
Given
radi
3.
statements asons
oP LM
oc Lc
re
≅
≅ uus of a circle
of a circlePc Mc radiuscPo cML
≅≅ ssss
oX LYocX LcY
Xc Yc
≅≅
≅
Definition of BisectorHL
ccPctrcB
c
D
F
a
e
Lesson 261.
statements asons
Xc Yc radius
re
≅ of a cirrclea tangent of a circle
is perpendic∠ ≅ ∠PYc PXc uular to the
radius at that point.
Pc Pc flexiv≅ re eePYc PXc HL
PX PY cPctrc
statements asons
De a
≅
≅
⊥
2.
re
BB
ac Bc radius
Fc Fc flexiveFc
Given
of a circle≅
≅ re aa FcB HL
ace Bce cPctrc
ae Be
≅∠ ≅ ∠
≅
Property of
centraal angle
Given
radi
3.
statements asons
oP LM
oc Lc
re
≅
≅ uus of a circle
of a circlePc Mc radiuscPo cML
≅≅ ssss
oX LYocX LcY
Xc Yc
≅≅
≅
Definition of BisectorHL
ccPctrc
Y
M
L
P
c
X
o
Honors Lesson 28Lesson 281. 67 1
2134
50 12
100
180 50 6
=
=
=
=
− +
( )
º
( )
º
X
X
Y
Y
77 12
63 12
126
180 77 103
180 84
( ) =
=
=
= − =
= −
( )
( )
º
º
Z
Z
Z
B
a
2.
==
= =
= ( ) =
∠ = =
96
2 77 154
2 63 126
12
º
º
º º
c x
mQr
m Qcr mQr
3.
66
40 302
702
35
35
11
º
º º º º
º
4.
5.
m aec
m BeD
m KPL
∠ = + = =
∠ =
∠ = 66 362
802
40º º º º− = =
Honors Lesson 29Lesson 291.
angle tan
º .º .º .ºº .
10 1815 26830 5845 160 1 773
1020
1820
18 20
3 6
2. tan º
.
.
.
=
=
= ( )=
a
a
a
a mi or 19,0008 ft
ft = a
3. tan º45150
1150
150
=
=
a
a
Honors Lesson 29 - Honors Lesson 30
soLUtions GeoMetrY224
Lesson 291.
angle tan
º .º .º .ºº .
10 1815 26830 5845 160 1 773
1020
1820
18 20
3 6
2. tan º
.
.
.
=
=
= ( )=
a
a
a
a mi or 19,0008 ft
ft = a
3. tan º45150
1150
150
=
=
a
a
a 10º
20 mi
Lesson 291.
angle tan
º .º .º .ºº .
10 1815 26830 5845 160 1 773
1020
1820
18 20
3 6
2. tan º
.
.
.
=
=
= ( )=
a
a
a
a mi or 19,0008 ft
ft = a
3. tan º45150
1150
150
=
=
a
a a 45º 150 ft
Honors Lesson 30Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L
55º
14 w
Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L
h 3
Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L
4.5
34º
p
Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L
55º
14
g
Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L
500 7º
g
Lesson 301. sin º
.
.
.
5514
819214
14 8192
11
=
=
( ) =
=
w
w
w
w 44688
43
06983
3 0698
2094
ft
mi
2. sin º
.
.
.
=
=
( ) =
=
h
h
h
h
..2094 5,280 ft
( ) =
=
=
1 105 632
344 5
5592
, .
sin º.
.
3. p
p44 5
4 5 5592
2 5164
.. .
.
c
( ) =
=
p
p mi or 13,286.592 ft
4. oos º
.
.
.
cos
ft
5514
573614
14 5736
8 0304
=
=
( ) =
=
g
g
g
g
5. 7
ft
º
.
.
.
cos
=
=
( ) =
=
g
g
g
g
500
9925500
500 9925
496 25
6. 60
ft
º
.
.
=
=
=
=
30
5 30
5 30
60
L
LL
L60º
L
30
test 1 - test 4GeoMetrY 225
Test 1Test11.
2.
3.
4.
B
e
a
a
:
:
:
:
no dimensions
line
point
lenggth
is infinite
point e
point Q
5.
6.
7.
8.
B
c
a
D
:
:
:
: ° iinfinite
line segment aB
ray ce
p
( )9.
10.
11.
e
B
c
:
:
: ooint
equal
similar
congruent
a: r
12.
13.
14.
15.
c
a
B
:
:
:
aay
Test 2Test 21.
2.
3.
4
e
D
c
:
:
:
coplanar
line
length and width
..
5.
6.
7.
8.
c two
a
B union
B
c
:
:
:
:
intersection
infinite
::
:
∅ ( )∩ ( )
null or empty set
intersection9.
10.
a
e ::
:
:
:
∪ ( )union
line
points
11.
12.
13.
B line cD
c aH
e FF, D, e
point e
14.
15.
c line eG
a
:
:
Test 3Test 31.
2.
3.
4.
B
B
c
a
:
:
:
:
point H
point H
point J
pointt H
aHc
protractor
degrees
srt
5.
6.
7.
8.
9.
e
B
c
B
:
:
:
:
∠
∠
ee
B
a
D
:
: º
: º
the measure of angle two
10.
11.
12.
45
90
:: º
: º
: º
: º
80
130
25
160
13.
14.
15.
e
a
e
Test 31.
2.
3.
4.
B
B
c
a
:
:
:
:
point H
point H
point J
pointt H
aHc
protractor
degrees
srt
5.
6.
7.
8.
9.
e
B
c
B
:
:
:
:
∠
∠
ee
B
a
D
:
: º
: º
the measure of angle two
10.
11.
12.
45
90
:: º
: º
: º
: º
80
130
25
160
13.
14.
15.
e
a
e
Test 4Test 41.
2.
3.
4.
5.
D BGe
c cGe
a aGc
c
e
:
:
:
:
:
∠
∠
∠
reflex
straaight
point s
can't tell from inf
6.
7.
8.
B
B
e
:
: º
:
90
oormation
given; it may be anything
between 90º and 180º
acute9.
10.
11.
12.
a
D
c rnP
e
:
: º
:
: º
90
30 45
∠
+ ºº º
: º º º
:
=
− =
75
90 50 4013.
14.
B
B of angles given, onlyy obtuse
ones are and
is the sma
∠ ∠
∠
MnQ Pns
Pns
.
lller of the two.
15. c m Mnr: º º º
º º
∠ = + =
+ =
60 30 90
90 91 1181º
reflex angles are > 180º
Test Solutions
test soLUtions
test 5 - test 8
soLUtions GeoMetrY226
Test 5Test 51.
2.
3.
a parallel
B
e
:
:
:
perpendicular
perpendiccular
bisector
i,
4.
5.
6.
7.
B
a aF FB
D Da and GF
c
:
:
:
:
=
iii and iV are true
8.
9.
10
B
B
: º º
: º º
90 2 45
90 2 45
÷
÷
=
=
..
11.
12.
c
a
a
:
:
:
⊥
this is the converse of
the origiinal statement.
i and iii:
straightedge an
13. c :
dd compass
at the vertex
perpendicular
14.
15.
D
c
:
: llines are
not parallel
Test 51.
2.
3.
a parallel
B
e
:
:
:
perpendicular
perpendiccular
bisector
i,
4.
5.
6.
7.
B
a aF FB
D Da and GF
c
:
:
:
:
=
iii and iV are true
8.
9.
10
B
B
: º º
: º º
90 2 45
90 2 45
÷
÷
=
=
..
11.
12.
c
a
a
:
:
:
⊥
this is the converse of
the origiinal statement.
i and iii:
straightedge an
13. c :
dd compass
at the vertex
perpendicular
14.
15.
D
c
:
: llines are
not parallel
Test 6Test 61.
2.
3.
e
c
B
:
:
: º º
supplementary
congruent
90 35− ==
− =
+ =
55
180 40 140
20 70 90
º
: º º º
: º º º
4.
5.
c
e , so
they arre complementary
, because
6.
7.
B and
a
:
: º
∠ ∠2 5
90 lline sV
can't tell from
information
⊥ line Wt
e8. :
given
D:
they combine to
form a s
9.
10.
∠1
180a : º
ttraight angle.
vertical angles
D: We don'
11.
12.
c :
tt know the measures
of 4 and 5, so sum canno∠ ∠ tt
be determined.
is a straight line, so13. a Fc: 1 would
be included to make 180º.
the me
∠↔
14. D : aasures of these angles
are not given: looking tthe
same is not sufficient.
15. a : º º º90 90 185+ <
Test 61.
2.
3.
e
c
B
:
:
: º º
supplementary
congruent
90 35− ==
− =
+ =
55
180 40 140
20 70 90
º
: º º º
: º º º
4.
5.
c
e , so
they arre complementary
, because
6.
7.
B and
a
:
: º
∠ ∠2 5
90 lline sV
can't tell from
information
⊥ line Wt
e8. :
given
D:
they combine to
form a s
9.
10.
∠1
180a : º
ttraight angle.
vertical angles
D: We don'
11.
12.
c :
tt know the measures
of 4 and 5, so sum canno∠ ∠ tt
be determined.
is a straight line, so13. a Fc: 1 would
be included to make 180º.
the me
∠↔
14. D : aasures of these angles
are not given: looking tthe
same is not sufficient.
15. a : º º º90 90 185+ <
Test 7Test 71.
2.
3.
D
c
e
:
: º º º
:
∠
− =
7
180 80 100
alternate interrior angles
are congruent.
alternate
4.
5.
B
D
:
:
∠2
eexterior angles
e 1, 2, 4, 5, 6, 7 and 86.
7
: '∠ s
..
8.
9.
c
D
e
: º;
:
:
65 vertical angles
vertical angles
ssupplementary angles
can't tell: rules for10. e : alternate
exterior angles apply only for
paralllel lines
if the sum of two angles is
180
11. c :
ºº, they are supplementary.
parallel lines12.
1
a :
33.
14.
15.
D
D
B
:
:
:
45º
: four for each intersection8
ccongruent
Test 8Test 81.
2.
e
c
:
:
i, ii and V
all squares have 4 righht
angles and opposite sides
that are congruentt, so
they are rectangles.
some trapezoids h3. D : aave 1
right angle, but they need
not have any.
44.
5.
6.
e
a quadrilateral
D
:
:
:
length of each side
1800
360
º
:
:
: º
:
7.
8.
9.
10.
1
D
B
a
B
square
rhombus
trapezoid
11.
12.
13.
a in
c m
D
:
:
:
5 7 9 3 24
9 10 15 34
+ + + =
+ + =
unlabeleed horizontal side
has a length of 8 5 4 4 5. .− = inn
P in
B P cm
e
= + + + + + =
= ( ) =
4 3 4 5 2 8 5 5 27
4 11 44
. .
:14.
15. :: ftP = ( ) + ( ) = + =2 25 2 15 50 30 80
test 8 - Unit test i
soLUtionsGeoMetrY 227
Test 81.
2.
e
c
:
:
i, ii and V
all squares have 4 righht
angles and opposite sides
that are congruentt, so
they are rectangles.
some trapezoids h3. D : aave 1
right angle, but they need
not have any.
44.
5.
6.
e
a quadrilateral
D
:
:
:
length of each side
1800
360
º
:
:
: º
:
7.
8.
9.
10.
1
D
B
a
B
square
rhombus
trapezoid
11.
12.
13.
a in
c m
D
:
:
:
5 7 9 3 24
9 10 15 34
+ + + =
+ + =
unlabeleed horizontal side
has a length of 8 5 4 4 5. .− = inn
P in
B P cm
e
= + + + + + =
= ( ) =
4 3 4 5 2 8 5 5 27
4 11 44
. .
:14.
15. :: ftP = ( ) + ( ) = + =2 25 2 15 50 30 80
Test 81.
2.
e
c
:
:
i, ii and V
all squares have 4 righht
angles and opposite sides
that are congruentt, so
they are rectangles.
some trapezoids h3. D : aave 1
right angle, but they need
not have any.
44.
5.
6.
e
a quadrilateral
D
:
:
:
length of each side
1800
360
º
:
:
: º
:
7.
8.
9.
10.
1
D
B
a
B
square
rhombus
trapezoid
11.
12.
13.
a in
c m
D
:
:
:
5 7 9 3 24
9 10 15 34
+ + + =
+ + =
unlabeleed horizontal side
has a length of 8 5 4 4 5. .− = inn
P in
B P cm
e
= + + + + + =
= ( ) =
4 3 4 5 2 8 5 5 27
4 11 44
. .
:14.
15. :: ftP = ( ) + ( ) = + =2 25 2 15 50 30 80
Test 9Test 91.
2.
3.
B height
e
:
: perpendicular to the base
BB
D
a
:
:
:
divide by two
find the average base4.
5. 90ºº
:
, because they are
perpendicular
100 ft : 6. c 2 aarea is always
in square units
not enough i7. e : nnformation;
need to know both bases
8. D a bh: = =12
122
8 4 16 2
15 3 45 2
( ) ( ) =
= = ( ) ( ) =:
:
m
c a bh units
e
9.
10. nott enough information;
perpendicular height is neeeded
11.
12.
a a in
a a b
: . .
:
= + ( ) =
=
5 92
3 5 24 5 2
12
hh m
c a
= ( ) ( ) =
= ( ) ( ) + ( ) ( ) =
+ =
12
15 6 45 2
4 3 2 8 5
12 17
: .13.
229 2
11 10 110 2
ft
:14. e a bh cm= = ( ) ( ) =
all 4 sides of a rhombus
are congruent
15. a a bh: = = ( ) ( ) =25 15 375 ft2
Test 10Test101.
2.
3.
4.
D
c
B acute
c
:
:
:
:
obtuse
isosceles
if itt has a 90º angle,
the remaining angles
mu
two
sst add to 90º.
scalene
impos
90 28 62º º º
:
:
− =
5.
6.
B
e ssible to draw
acute and equ
61 62 61 184º º º º
:
+ + =
7. B iilateral
is the smallest number
among the
8. c : 6
choices which,
when added to 7 yields
a result greater than 12.
impossible to draw9. e :
;2 2 4+ = 44 5<
10.
11.
a
c
:
:
isosceles and right
equilateral, beecause the third
angle must also be 60º
12. D ri: gght
a
º º º
º º º
:
180 74 16
180 90 90
− +( ) =
− =
13. isosceles aand acute
equilateral14.
15.
B
a
:
: º º º º34 73 73 180+ + =
Test101.
2.
3.
4.
D
c
B acute
c
:
:
:
:
obtuse
isosceles
if itt has a 90º angle,
the remaining angles
mu
two
sst add to 90º.
scalene
impos
90 28 62º º º
:
:
− =
5.
6.
B
e ssible to draw
acute and equ
61 62 61 184º º º º
:
+ + =
7. B iilateral
is the smallest number
among the
8. c : 6
choices which,
when added to 7 yields
a result greater than 12.
impossible to draw9. e :
;2 2 4+ = 44 5<
10.
11.
a
c
:
:
isosceles and right
equilateral, beecause the third
angle must also be 60º
12. D ri: gght
a
º º º
º º º
:
180 74 16
180 90 90
− +( ) =
− =
13. isosceles aand acute
equilateral14.
15.
B
a
:
: º º º º34 73 73 180+ + =
Unit Test IUnit Test II
1.
2.
3.
intersection
congruent
empty or null
triangle
supplementary
reflex
bisect
4.
5.
6.
7. oor
angle8.
II1.
2.
check with protractor
check with protractorr: smaller
angles should each measure 40º
Unit test i - test 11
soLUtions GeoMetrY228
III1.
2.
a triangle with two equal sides
a trianglee with no equal sides
IV P ft= + + + =4 9 5 11 29
a ft= + ( ) = ( ) =11 92
3 10 3 30 2
V1.
2.
110
11 3 110
º:
º
vertical angles
correspo
m m∠ = ∠ =
nnding angles
supplem
m m∠ = − ∠ =
− =
9 180 11
180 110 70
º
º º º
eentary angles
1 & 9, 3 & 11, 2 & 10, 4 & 12, 3.
55 & 13, 7 & 15, 6 & 14, or 8 & 16
4 & 9, 3 & 4. 110, 7 & 14, or 8 & 13
, 4, 9, or 12: 's 5,5. ∠ ∠1 8, 13 and
16 appear to be acute but we
don't knnow for certain, because
no information is giveen about
these angles.
: ac is not parallel 6. no tto BD
B and c, or a and D
point a
infinit
↔ ↔
7.
8.
9. ee: only two points are
labeled, but every line contains
an infinite number of points.
one:10. length
11.
12.
∠ ∠ ∠ ∠
∠
2 3 10 11
9
, , , or
VIVII
a bh in= = ( ) ( ) =12
12
6 2 6 2
rectangle, square,
parrallelogram, rhombus
2, 3, 4, 5, 6 all
VIII1. { } : elements that
appear in either of the two setss
no: the element 2 is found
in set a, but n
2.
oot in set B
Test 11 Test111.
2.
c
a
:
:
4 diagonals,
forming 5 triangles
110 triangles
3. D n: º
º
−( ) =>
( ) −( ) = ( )2 180
11 2 180 9 180ºº , º
:
: º
: º
=
−( ) => ( )
1 620
360
2 180 8
4.
5.
6.
a pentagon
a
B n −−( ) =
( ) =
=
2 180
6 180 1 080
1 080 8 135
º
º , º
, º º
total
per÷ angle
the exterior angles of a
polygon alw
7. e :
aays add up to 360º.
8. c n: º º−( ) => ( ) −( ) =2 180 5 2 180
33 180 540
540 5 108
180
( ) =
=
∠ =
º º
º º
total
per angle
m b
÷
ºº º º
º º º
: º
− +( ) =
− =
−( ) => ( ) −
36 108
180 144 36
2 180 8 29. e n (( ) =
( ) =
=
∠ =
180
6 180 1 080
1 080 8 135
13
º
º , º
, º º
total
÷
m a 55 2 67 5
2 180
6 2 180 4 180
º . º
: º
º º
÷ =
−( ) =>
( ) −( ) = ( ) =
10. B n
7720
360 6 60
720
º
: º º
: º
11.
12.
a
c
÷ =
for all interior angles
from #10
for ea
( )=
=
720 6 120
120 2 60
º º
º º
÷
÷
cch new angle
13. D m QVr: º º º
º
∠ = − +( ) =
−
180 30 120
180 1500 30
120 30 90
º º
:
º º º
:
=
∠ = ∠ − ∠ =
− =
14.
15.
B m rVU m QVU m QVr
a mm trU
m srQ
∠ =
∠ = =÷ ÷4 120 4 30º º
test 12 - test 14
soLUtionsGeoMetrY 229
Test 12Test121.
2.
3.
a
B
B
:
: º º º
:
circumference
di
360 50 310− =
aameter
tangent
they are perpendicular
4.
5.
6.
B
c
e
:
:
::
:
:
:
:
inscribed in
sector
secant
arc
t
7.
8.
9.
10.
D
a
c
D hhe measure of an inscribed
angle is half that oof the
intercepted arc.
48 2 24º º÷ =
11.
12.
13.
14
c
B
D
..
15.
a
e
Test121.
2.
3.
a
B
B
:
: º º º
:
circumference
di
360 50 310− =
aameter
tangent
they are perpendicular
4.
5.
6.
B
c
e
:
:
::
:
:
:
:
inscribed in
sector
secant
arc
t
7.
8.
9.
10.
D
a
c
D hhe measure of an inscribed
angle is half that oof the
intercepted arc.
48 2 24º º÷ =
11.
12.
13.
14
c
B
D
..
15.
a
e
Test 13Test131.
2.
3.
4.
B
a
a r
B
:
:
:
:
radius
circumference
π
π
2
2 rr
e
B
e
5.
6.
7.
:
: '
:
12
12
60
long axis short axis
l
⋅ ⋅ π
aatitude
the prime meridian8.
9.
c
c a r
:
: .= ( )π ≈2 3 14 322
28 26 2
2 2 3 14 3
18 84
( ) =
= ( ) ( ) ( ) =
.
: .
.
units
10. B c rπ ≈
uunits
a
c
radius is half the diameter( )
11.
12.
: 227
::
( )
:
a r
in
B c
=
( ) = × =
=
π ≈2
227
72 22
7
497
1
154 2
13. 22 21
22
7
71
441
44
π ≈r
in
( )( )( ) =
=
144.
15.
B a
in
c a r
: .
.
:
= ( ) ( )( ) ( ) ( ) =
=
5 2 10 3 14
31 4 2
2
π ≈
π ≈ 33 14 42
50 24 2
.
.
( ) ( ) =
units
Test131.
2.
3.
4.
B
a
a r
B
:
:
:
:
radius
circumference
π
π
2
2 rr
e
B
e
5.
6.
7.
:
: '
:
12
12
60
long axis short axis
l
⋅ ⋅ π
aatitude
the prime meridian8.
9.
c
c a r
:
: .= ( )π ≈2 3 14 322
28 26 2
2 2 3 14 3
18 84
( ) =
= ( ) ( ) ( ) =
.
: .
.
units
10. B c rπ ≈
uunits
a
c
radius is half the diameter( )
11.
12.
: 227
::
( )
:
a r
in
B c
=
( ) = × =
=
π ≈2
227
72 22
7
497
1
154 2
13. 22 21
22
7
71
441
44
π ≈r
in
( )( )( ) =
=
144.
15.
B a
in
c a r
: .
.
:
= ( ) ( )( ) ( ) ( ) =
=
5 2 10 3 14
31 4 2
2
π ≈
π ≈ 33 14 42
50 24 2
.
.
( ) ( ) =
units
Test 14Test141.
2.
B
e
:
:
the area of the base
all of the abbove
edges
8 vertices
cubic units
oft
3.
4.
5.
e
c
D
:
:
:
een written as units3( )×
= ( ) ( ) ( )6.
7.
a r h
B V
:
:
π 2
6 6 6 ==
= ( ) ( ) ( ) =
216 3
3 4 9 108 3
:
:
in
D
c V units
8.
9.
10.
faces
ee V Bh r h
m
a face
:
. ,
:
= =
( ) ( ) ( ) =
π ≈2
3 14 102 6 1 884 3
611. ss
e V m
B V
12.
13.
:
: ,
= ( ) ( ) ( ) =
= ( ) ( ) ( ) =
5 8 2 80 3
10 10 10 1 0000 3
2
3 14 102 16 5 024 3
:
. , ft
in
B V Bh r h14. = =
( ) ( ) ( ) =
π ≈
115. c V Bh r h: .
ft
= = ( ) ( ) ( ) =π ≈2 3 14 52 10
785 3
test 15 - test 16
soLUtions GeoMetrY230
Test 15Test151.
2.
3.
a
B
a
:
:
:
triangles
altitude
slant heightt
prism
cylinder:
4.
5.
D
a
V Bh r h
:
:
. .
= =
( ) ( )π ≈2
3 14 1 52 4(( ) =
= =
( ) ( )( ) =
28 26
13
13
2
13
3 14 1 52 4 9
.
. .
cone:
V Bh r hπ ≈
..
:
42
6. D not enough information,
because we do noot know
the heights
sphere:7. a
V r
:
.= ( )43
3 43
3 14π ≈ 443 267 95
13
13
2
13
3 14 42 4
( )
= =
( ) ( )( )
≈
π ≈
.
.
cone: V Bh r h
≈≈ 66 99.
Test151.
2.
3.
a
B
a
:
:
:
triangles
altitude
slant heightt
prism
cylinder:
4.
5.
D
a
V Bh r h
:
:
. .
= =
( ) ( )π ≈2
3 14 1 52 4(( ) =
= =
( ) ( )( ) =
28 26
13
13
2
13
3 14 1 52 4 9
.
. .
cone:
V Bh r hπ ≈
..
:
42
6. D not enough information,
because we do noot know
the heights
sphere:7. a
V r
:
.= ( )43
3 43
3 14π ≈ 443 267 95
13
13
2
13
3 14 42 4
( )
= =
( ) ( )( )
≈
π ≈
.
.
cone: V Bh r h
≈≈ 66 99.
8. c : same, because the spheres
are the same size
99. B V Bh: cylinder:
rectangular solid
= = ( ) ( ) =4 10 40
::
V Bh= = ( ) ( ) =
>
10 8 80
80 40
10. e
V r
: none of the above
correct formula is = 43
π 33
1313
13
6 6 10
120 3
11.
12.
13
B V Bh
a V Bh
in
:
:
=
= = ( ) ( ) ( ) =
..
14.
e V Bh r h
in
a
:
.
:
= =
( ) ( )( ) =
13
13
2
13
3 14 52 12 314 3
π ≈
VV Bh
B V r
= = ( ) ( ) ( ) =
= ( )
12
3 4 6 36 3
43
3 43
3 14 63
ft
: .15. π ≈ (( ) =
904 32 3. m
10. e
V r
: none of the above
correct formula is = 43
π 33
1313
13
6 6 10
120 3
11.
12.
13
B V Bh
a V Bh
in
:
:
=
= = ( ) ( ) ( ) =
..
14.
e V Bh r h
in
a
:
.
:
= =
( ) ( )( ) =
13
13
2
13
3 14 52 12 314 3
π ≈
VV Bh
B V r
= = ( ) ( ) ( ) =
= ( )
12
3 4 6 36 3
43
3 43
3 14 63
ft
: .15. π ≈ (( ) =
904 32 3. m
Test 16Test161.
2.
3.
4.
D
c
B
c
:
:
:
:
6 faces
5 faces
4 faces
two circles and the rectangle
formed by "unrollingg" the side
square units
6 7
5.
6.
D
a in
:
: ( )( ) =7 294 2
77. a sa: =
( )( ) + ( ) ( ) + ( ) ( ) =
+ +
2 12 14 2 12 8 2 8 14
336 192 224 ==
= +
( ) ( ) ( ) + ( )
752 2
2 2 2
2 3 14 52 2 3 14
ft
:
. .
8. a sa r rhπ π ≈
(( ) ( ) ( ) =
+ =
= ( ) ( )
5 10
157 314 471 2
4 12
6 9: (
m
D sa9. + ( ) ( ) =
+ =
= ( ) ( ) + (
)
:
6 6
108 36 144 2
2 3 4 2 4
units
B sa10. )) ( ) + ( ) ( ) =
+ + =
6 2 3 6
24 48 36 108 2units
11. B : since the square base has an
area of 100 ftt , it must be
100 or 10 ft on a side.
2
4 12
1sa = ( 00 20 10 10
400 100 500 2
( ) ( ) + ( ) ( ) =
+ =
)
ft
:12. a sa == ( ) ( ) + ( ) ( ) =
+ =
4 12
20 30 20 20
1 200 400 1 600
( )
, , mm2
13. D sa:
ft
= ( ) ( ) + ( )( ) + ( ) ( ) =
+ + =
2 1 3 2 1 4 2 3 4
6 8 24 38 2
test 16 - test 18
soLUtionsGeoMetrY 231
14. c sa r rh:
. .
.
= +
( ) ( ) + ( ) ( ) ( ) =
2 2 2
2 3 14 32 2 3 14 3 5
56
π π ≈
552 94 2 150 72 2
2 7 5 70
+ =
= ( ) ( ) =
. .
: " " :
cm
e roof m15. sa 22
2 12
6 4 24 2
2 2 7
triangles:
sa m
sides
sa
= ( ) ( ) =
= ( )(:
)) + ( ) ( ) =
= ( ) ( ) =
2 2 6 52 2
6 7 42 2:
:
m
bottom sa m
total
saa m= + + + =70 24 52 42 188 2
Test 17Test171.
2.
3.
4.
c
B
B rs
e r s r s
:
:
:
:
a whole number
6
+ = + ::
cannot be simplified
5.
6.
B X Y XY
D
:
:
5 6 30
10 3
( )( ) =
77.
8.
9.
a
B
c
:
:
:
3 5 3 5 9 25 9 5 45
45 9 5 3 5
24 4 6
( )( ) = = ( ) =
= =
= = 22 6
42 42
24 18
6 9
10.
11.
e
a
: :
:
=
=
cannot be simplified
224 26
4 21
4 2
15 8
5 2
15 45
3 41
3 4 3 2 6
= =
= = =
= ( ) =
12.
13.
c
e
:
: caannot be simplified
14.
1
B : 2 3 3 3 6 3
2 3 6 3 11 3
+ + =
+ +( ) =
55. D : 5 3 4 2 20 6( )( ) =
Test171.
2.
3.
4.
c
B
B rs
e r s r s
:
:
:
:
a whole number
6
+ = + ::
cannot be simplified
5.
6.
B X Y XY
D
:
:
5 6 30
10 3
( )( ) =
77.
8.
9.
a
B
c
:
:
:
3 5 3 5 9 25 9 5 45
45 9 5 3 5
24 4 6
( )( ) = = ( ) =
= =
= = 22 6
42 42
24 18
6 9
10.
11.
e
a
: :
:
=
=
cannot be simplified
224 26
4 21
4 2
15 8
5 2
15 45
3 41
3 4 3 2 6
= =
= = =
= ( ) =
12.
13.
c
e
:
: caannot be simplified
14.
1
B : 2 3 3 3 6 3
2 3 6 3 11 3
+ + =
+ +( ) =
55. D : 5 3 4 2 20 6( )( ) =
Test 18Test181.
2.
D a b c
c
:
:
2 2 2+ =
the triangle is a right ttriangle
of the
3. D H
H
H
H
:
.
32 22 2
9 4 2
13 2
13 3 61
+ =
+ =
=
= ≈
answers given,
4 is closest.
4. c a B H
a B
: 2 2 2
2 2
+ =
+ ==
+ =
+ =
+ =
=
=
+
H
a B H
B H
H
H
H
e
2
2 2
32 72 2
9 49 2
58 2
58
42 6
5.
6.
:
: 22 2
16 36 2
52 2
52 4 13 2 13
62 82 102
36 6
=
+ =
=
= = =
+ =+
H
H
H
H
c7. :44 100
100 100== : true
since the Pythagorean
theoremm applies to this
triangle, it is a right trianngle.
not true
si
8. B :
:
52 92 122
25 81 144106 144
+ =+ =
=
nnce the Pythagorean
theorem does not apply to
tthis triangle, it is not a
right triangle.
9. c : 900
52 2 612
25 2 612 36
º
:
:
10.
11.
a
D L
L
L
hypotenuse
+ = ( )+ =
=LL
B L
L
LL
a st
=
+ =
+ =
==
=
6
122 2 132
144 2 1692 25
5
12
12.
13.
:
:
1122 2 202
144 2 4002 256
16
12
12
24
+ =
+ =
==
= =
L
L
LL
B a bh14. : (( ) ( )
=
= ( ) =
16
192 2
8 192 1 536 2: ,
units
e a units15.
Test181.
2.
D a b c
c
:
:
2 2 2+ =
the triangle is a right ttriangle
of the
3. D H
H
H
H
:
.
32 22 2
9 4 2
13 2
13 3 61
+ =
+ =
=
= ≈
answers given,
4 is closest.
4. c a B H
a B
: 2 2 2
2 2
+ =
+ ==
+ =
+ =
+ =
=
=
+
H
a B H
B H
H
H
H
e
2
2 2
32 72 2
9 49 2
58 2
58
42 6
5.
6.
:
: 22 2
16 36 2
52 2
52 4 13 2 13
62 82 102
36 6
=
+ =
=
= = =
+ =+
H
H
H
H
c7. :44 100
100 100== : true
since the Pythagorean
theoremm applies to this
triangle, it is a right trianngle.
not true
si
8. B :
:
52 92 122
25 81 144106 144
+ =+ =
=
nnce the Pythagorean
theorem does not apply to
tthis triangle, it is not a
right triangle.
9. c : 900
52 2 612
25 2 612 36
º
:
:
10.
11.
a
D L
L
L
hypotenuse
+ = ( )+ =
=LL
B L
L
LL
a st
=
+ =
+ =
==
=
6
122 2 132
144 2 1692 25
5
12
12.
13.
:
:
1122 2 202
144 2 4002 256
16
12
12
24
+ =
+ =
==
= =
L
L
LL
B a bh14. : (( ) ( )
=
= ( ) =
16
192 2
8 192 1 536 2: ,
units
e a units15.
test 18 - test 19
soLUtions GeoMetrY232
Test181.
2.
D a b c
c
:
:
2 2 2+ =
the triangle is a right ttriangle
of the
3. D H
H
H
H
:
.
32 22 2
9 4 2
13 2
13 3 61
+ =
+ =
=
= ≈
answers given,
4 is closest.
4. c a B H
a B
: 2 2 2
2 2
+ =
+ ==
+ =
+ =
+ =
=
=
+
H
a B H
B H
H
H
H
e
2
2 2
32 72 2
9 49 2
58 2
58
42 6
5.
6.
:
: 22 2
16 36 2
52 2
52 4 13 2 13
62 82 102
36 6
=
+ =
=
= = =
+ =+
H
H
H
H
c7. :44 100
100 100== : true
since the Pythagorean
theoremm applies to this
triangle, it is a right trianngle.
not true
si
8. B :
:
52 92 122
25 81 144106 144
+ =+ =
=
nnce the Pythagorean
theorem does not apply to
tthis triangle, it is not a
right triangle.
9. c : 900
52 2 612
25 2 612 36
º
:
:
10.
11.
a
D L
L
L
hypotenuse
+ = ( )+ =
=LL
B L
L
LL
a st
=
+ =
+ =
==
=
6
122 2 132
144 2 1692 25
5
12
12.
13.
:
:
1122 2 202
144 2 4002 256
16
12
12
24
+ =
+ =
==
= =
L
L
LL
B a bh14. : (( ) ( )
=
= ( ) =
16
192 2
8 192 1 536 2: ,
units
e a units15.
Test 19Test191.
2.
3.
B
e
a
:
:
:
denominator
common denominat
1
oor
4.
5.
6.
a
D
B
:
:
:
5
3
5 3
3 3
5 3
9
5 33
8 2
4
8 22
4 21
4 2
4 3
8
= = =
= = =
== = =
= =
= =
=
4 3 2
8 2
4 6
16
4 64
61
6
5 5
5
51
5
3 7
10
3 7 10
1
7.
8.
B
a
:
:00 10
3 70
100
3 7010
4 15
6 6
4 15 6
6 6 6
4 90
6 36
4 906 6
=
=
= = =9. c :
(( ) = = =
= =
= =
4 9036
909
9 109
3 109
103
15 11
5
15 11 5
5 5
110. D : 55 55
25
15 555
3 551
3 55
4 3
2
2 3
2
6 3
2
6 3 2
2 2
6 6
4
=
= =
+ = = =11. c :
== = =
+ = + =
+ = +
6 62
3 61
3 6
7
5
3
2
7 5
5 5
3 2
2 2
7 5
25
3 2
4
7 55
3
12. a :
222
7 5 25 2
3 2 52 5
14 510
15 210
14 5 15 210
=
( )( ) +
( )( ) =
+ = +
13.. D : 8 6
3
5 3
2
8 6 3
3 3
5 3 2
2 2
8 18
9
5 6
4
8 183
5 62
8 9 23
− = − =
− = − =
− 55 62
8 3 23
5 62
8 21
5 62
8 2 21 2
5 62
16 22
5 62
=( )
− =
− =( )
( ) − =
− = 116 2 5 62
6 11
3
2 5
2
6 11 3
3 3
2 5 2
2 2
6 33
9
2 10
4
6 3
−
− = − =
− =
14. e :
333
2 102
2 331
101
2 33 10
2 2
8
7 3
3
2
4
71
22
7
− =
− = −
+ = + =
+
15. e :
== + =1 7 8
Test191.
2.
3.
B
e
a
:
:
:
denominator
common denominat
1
oor
4.
5.
6.
a
D
B
:
:
:
5
3
5 3
3 3
5 3
9
5 33
8 2
4
8 22
4 21
4 2
4 3
8
= = =
= = =
== = =
= =
= =
=
4 3 2
8 2
4 6
16
4 64
61
6
5 5
5
51
5
3 7
10
3 7 10
1
7.
8.
B
a
:
:00 10
3 70
100
3 7010
4 15
6 6
4 15 6
6 6 6
4 90
6 36
4 906 6
=
=
= = =9. c :
(( ) = = =
= =
= =
4 9036
909
9 109
3 109
103
15 11
5
15 11 5
5 5
110. D : 55 55
25
15 555
3 551
3 55
4 3
2
2 3
2
6 3
2
6 3 2
2 2
6 6
4
=
= =
+ = = =11. c :
== = =
+ = + =
+ = +
6 62
3 61
3 6
7
5
3
2
7 5
5 5
3 2
2 2
7 5
25
3 2
4
7 55
3
12. a :
222
7 5 25 2
3 2 52 5
14 510
15 210
14 5 15 210
=
( )( ) +
( )( ) =
+ = +
13.. D : 8 6
3
5 3
2
8 6 3
3 3
5 3 2
2 2
8 18
9
5 6
4
8 183
5 62
8 9 23
− = − =
− = − =
− 55 62
8 3 23
5 62
8 21
5 62
8 2 21 2
5 62
16 22
5 62
=( )
− =
− =( )
( ) − =
− = 116 2 5 62
6 11
3
2 5
2
6 11 3
3 3
2 5 2
2 2
6 33
9
2 10
4
6 3
−
− = − =
− =
14. e :
333
2 102
2 331
101
2 33 10
2 2
8
7 3
3
2
4
71
22
7
− =
− = −
+ = + =
+
15. e :
== + =1 7 8
Unit test i i - Unit test i i
soLUtionsGeoMetrY 233
Unit Test IIUnit Test III
1.
2.
3.
4.
pentagon
hypotenuse
sector
priism
rhombus
chord
sphere
latitude
5.
6.
7.
8.
II V = ( )10 6(( ) ( ) =
=
4 240 3in
area areaIII of two circles plus
aarea of "unrolled" rectangle =
+
( )2 2 2
2 3 14
π π ≈r rh
. 552 2 3 14 5 6
157 188 4 345 4 2
( ) + ( ) ( ) ( ) =
+ =
.
. . in
IV1. 2 6 5 10 2 5 6 10
10 60 10 4 15 10 2 15
20 15
( )( ) = ( ) ( ) =
= = ( ) =
22.
3.
3 7 2 71 5 3
2 7 12
7 3
− +
− + −
:
cannot be simplified
3 722
7
3 2 12
32
7 1 22
7
1 1 7 0
=
− + − = + − =
+ −( )( ) = ( ) 77 0
3
6
1
2
1 2
2 2
2
4
22
=
= = = =4.
V1. n
total
−( ) =>
( ) −( ) = ( ) =
2 180
6 2 180 4 180
720
720
º
º º
º
ºº º
º
÷6 120
360
= per angle
: the sum of the exter2. iior
angles of a regular polygon is
always 360º..
VI
1. a r= =
=
π ≈2 227
727
1154
1154
( )( )
ft22
2 21
22
7
71
441
( )( )( )2. c r= =
=
π ≈
444 ft
VIIVIII
check with protractor
area of 4 triangularr faces:
a bh= =
( ) ( ) ( ) =
4 12
4 12
4 5 40
( )
( ) in
a in
2
4 4 16 2
40
area of base:
total area
= ( )( ) =
= +116 56 2= in
IX1. the measure of an intercepted
arc is the samme as the measure
of the central angle that
inttercepts it, so m aXc
the measure of an i
∠ = 82º
2. nnscribed
angle is half the measure of the
arc iit intercepts, so
m aBc∠ = =82 2 41º º÷
X Leg Leg Hypotenuse or
L L H or a B c
2 2 2
2 2 2 2 2 2
+ =
+ = + =
1..
2.
L
L
LL
L
L
2 62 102
2 36 1002 64
8
2 22 132
2
+ =
+ =
==
+ = ( )+
ft
44 132 9
3
2 22
5 22 2
2 2 2 2 5 5
=
==
( ) + ( ) =
( )( ) + ( )
LL units
H3.
(( ) =
+ =
( ) + ( ) =
+ =
=
2 2 2
4 4 5 4 2
4 2 25 2 2
8 50 2
58 2
58
H
H
H
H
H
unitts H=
Unit test i i - test 21
soLUtions GeoMetrY234
X Leg Leg Hypotenuse or
L L H or a B c
2 2 2
2 2 2 2 2 2
+ =
+ = + =
1..
2.
L
L
LL
L
L
2 62 102
2 36 1002 64
8
2 22 132
2
+ =
+ =
==
+ = ( )+
ft
44 132 9
3
2 22
5 22 2
2 2 2 2 5 5
=
==
( ) + ( ) =
( )( ) + ( )
LL units
H3.
(( ) =
+ =
( ) + ( ) =
+ =
=
2 2 2
4 4 5 4 2
4 2 25 2 2
8 50 2
58 2
58
H
H
H
H
H
unitts H=
4. ( ) ( )1
22 1
32 2
1 1
2 2
1 1
3 3
+ =
( )( )+
( )( )=
H
H22
1
4
1
92
12
13
2
36
26
2
56
2
+ =
+ =
+ =
=
H
H
H
H
56
5
6
5 6
6 6
30
36
306
=
=
=
=
=
H
H
H
H
units H
Test 20Test 201.
2.
3.
B
D
c
:
:
:
hypotenuse
congruent
isosceles
44.
5.
6.
7.
e
a
c
:
:
:
Pythagorean theorem
B: 2
25 2
3 2 2 3 4= = 33 2 6
9 2
2
91
9
2
2
2 2
2 2
2 2
4
2 22
( ) =
= =
=
= = = =
8.
9.
a
B
:
: one leg
221
2
2 2 2 2
7
=
= + =both legs
beca
10.
11.
e a B and c
a
: ,
: uuse it is a
triangle
and the legs a
45 45 90º º º− −
rre congruent
by rule for
tr
12. c :
º º º
7 2
45 45 90− − iiangles
13.
14
D m: º º º
º º º
∠ = − +( ) =
− =
α 180 90 45
180 135 45
..
15.
a
e
:
:
2 3
2 3 2 2
because the legs
are congruent
= 66 so none
of the above
Test 201.
2.
3.
B
D
c
:
:
:
hypotenuse
congruent
isosceles
44.
5.
6.
7.
e
a
c
:
:
:
Pythagorean theorem
B: 2
25 2
3 2 2 3 4= = 33 2 6
9 2
2
91
9
2
2
2 2
2 2
2 2
4
2 22
( ) =
= =
=
= = = =
8.
9.
a
B
:
: one leg
221
2
2 2 2 2
7
=
= + =both legs
beca
10.
11.
e a B and c
a
: ,
: uuse it is a
triangle
and the legs a
45 45 90º º º− −
rre congruent
by rule for
tr
12. c :
º º º
7 2
45 45 90− − iiangles
13.
14
D m: º º º
º º º
∠ = − +( ) =
− =
α 180 90 45
180 135 45
..
15.
a
e
:
:
2 3
2 3 2 2
because the legs
are congruent
= 66 so none
of the above
Test 201.
2.
3.
B
D
c
:
:
:
hypotenuse
congruent
isosceles
44.
5.
6.
7.
e
a
c
:
:
:
Pythagorean theorem
B: 2
25 2
3 2 2 3 4= = 33 2 6
9 2
2
91
9
2
2
2 2
2 2
2 2
4
2 22
( ) =
= =
=
= = = =
8.
9.
a
B
:
: one leg
221
2
2 2 2 2
7
=
= + =both legs
beca
10.
11.
e a B and c
a
: ,
: uuse it is a
triangle
and the legs a
45 45 90º º º− −
rre congruent
by rule for
tr
12. c :
º º º
7 2
45 45 90− − iiangles
13.
14
D m: º º º
º º º
∠ = − +( ) =
− =
α 180 90 45
180 135 45
..
15.
a
e
:
:
2 3
2 3 2 2
because the legs
are congruent
= 66 so none
of the above
Test 21Test 211.
2.
D
a
: º º º
º º º
:
180 60 30
180 90 90
− +( ) =
− =
scalenne
2 times as long
B: dividing by 2
ti
3.
4.
5.
D
c
:
: 3 mmes as long
the side opposite the
30º angle
6. B :
is the short side, so
the hypotenuse would be
22 4 5( ) =
( ) =
=
= =
8 5
2 2 4
14 2 7
12
3
12 3
3 3
7.
8.
9.
e a a
e r r
B
:
:
:
÷
112 3
9
12 33
4 31
4 3
180 90 60
180 1
=
= =
∠ =
− +( ) =
−
10. c m:
º º º
º
α
550 30
14 2 7
7 3
180 90 30
º º
:
:
:
º º º
=
=
∠ =
− +
11.
12.
13.
a
D
B m
÷
β
(( ) =
− =
( ) =
180 120 60
4 3
2 4 8
º º º
:
:
14.
15.
a
c
test 21 - test 24
soLUtionsGeoMetrY 235
Test 211.
2.
D
a
: º º º
º º º
:
180 60 30
180 90 90
− +( ) =
− =
scalenne
2 times as long
B: dividing by 2
ti
3.
4.
5.
D
c
:
: 3 mmes as long
the side opposite the
30º angle
6. B :
is the short side, so
the hypotenuse would be
22 4 5( ) =
( ) =
=
= =
8 5
2 2 4
14 2 7
12
3
12 3
3 3
7.
8.
9.
e a a
e r r
B
:
:
:
÷
112 3
9
12 33
4 31
4 3
180 90 60
180 1
=
= =
∠ =
− +( ) =
−
10. c m:
º º º
º
α
550 30
14 2 7
7 3
180 90 30
º º
:
:
:
º º º
=
=
∠ =
− +
11.
12.
13.
a
D
B m
÷
β
(( ) =
− =
( ) =
180 120 60
4 3
2 4 8
º º º
:
:
14.
15.
a
c
Test 211.
2.
D
a
: º º º
º º º
:
180 60 30
180 90 90
− +( ) =
− =
scalenne
2 times as long
B: dividing by 2
ti
3.
4.
5.
D
c
:
: 3 mmes as long
the side opposite the
30º angle
6. B :
is the short side, so
the hypotenuse would be
22 4 5( ) =
( ) =
=
= =
8 5
2 2 4
14 2 7
12
3
12 3
3 3
7.
8.
9.
e a a
e r r
B
:
:
:
÷
112 3
9
12 33
4 31
4 3
180 90 60
180 1
=
= =
∠ =
− +( ) =
−
10. c m:
º º º
º
α
550 30
14 2 7
7 3
180 90 30
º º
:
:
:
º º º
=
=
∠ =
− +
11.
12.
13.
a
D
B m
÷
β
(( ) =
− =
( ) =
180 120 60
4 3
2 4 8
º º º
:
:
14.
15.
a
c
Test 22Test 221.
2.
3.
B
e
:
:
unproven and obvious
postulates
cc
a
B
D
:
: º
:
:
congruent
rhombus
complementar
4.
5.
6.
360
yy
congruent
supplementary
tr
7.
8.
9.
10.
B
e
c
B
: º
:
:
:
180
aapezoid
the figure described may
11.
12.
c r s t
c
:
:
+ >
be a rhombus, rectangle
or square, but is defiinitely
a parallelogram.
right
perpendi
13.
14.
B
e
:
: ccular
parallel15. D :
Test 23Test 231.
2.
3.
4.
5.
B
D
B sV
D tVs
a
:
: º
:
:
:
congruent
180
∠
∠VVst
a sVt
e tV
B srQ
D m c
6.
7.
8.
9.
:
:
:
: º º º
∠
∠ = − =180 118 62
110.
11.
e m B
B m a
: º º º
: º º º
∠ = − =
∠ = − +( ) =
180 113 67
180 62 67
1180 129 51
180 51 129
18
º º º
: º º º
:
− =
∠ = − =
∠ =
12.
13.
B m D
B m B 00 110 70
180 45 70
180 115 6
º º º
: º º º
º º
− =
∠ = − +( ) =
− =
14. c m a
55
180 65 115
º
: º º º15. D m D∠ = − =
Test 231.
2.
3.
4.
5.
B
D
B sV
D tVs
a
:
: º
:
:
:
congruent
180
∠
∠VVst
a sVt
e tV
B srQ
D m c
6.
7.
8.
9.
:
:
:
: º º º
∠
∠ = − =180 118 62
110.
11.
e m B
B m a
: º º º
: º º º
∠ = − =
∠ = − +( ) =
180 113 67
180 62 67
1180 129 51
180 51 129
18
º º º
: º º º
:
− =
∠ = − =
∠ =
12.
13.
B m D
B m B 00 110 70
180 45 70
180 115 6
º º º
: º º º
º º
− =
∠ = − +( ) =
− =
14. c m a
55
180 65 115
º
: º º º15. D m D∠ = − =
Test 24Test 241.
2.
B
a
:
:
the other two angles
they are conggruent by sss
i, ii, and iV, because of sas3. e :
44.
5.
6.
7.
8.
9.
a a a
c GH He
D
c sss
B
a
:
:
:
:
:
=
≅
congruent
sas
::
:
definition of a rhombus
reflexive propert10. D yy
sss
definition of midpoint
verti
11.
12.
13.
c
D
c
:
:
: ccal angles are congruent
B: sas
postulate
14.
15. B : ss are unproven
tatements used to prove
theore
s
mms
test 25 - test 28
soLUtions GeoMetrY236
Test 25Test 251. e aaa: , because sides may be
different leengths
hey are congruent.
prove the tri
2.
3.
D t
a
:
: aangles congruent
the midpoint
f one set
4.
5.
c
B i
:
: of corresponding
sides are congruent, the
triaangles may be proved
congruent by asa or aas.
6. aa JKZ XKZ
c
e asa
B
:
:
:
:
∠ ≅ ∠
7.
8.
9.
parallelogram
congruennt
reflexive property
definition of a
10.
11.
a
D
:
: bbisector
definition of a paralle
12.
13.
B rV tV
e
:
:
≅
llogram
alternate interior angles14.
15.
c
a asa
:
:
Test 251. e aaa: , because sides may be
different leengths
hey are congruent.
prove the tri
2.
3.
D t
a
:
: aangles congruent
the midpoint
f one set
4.
5.
c
B i
:
: of corresponding
sides are congruent, the
triaangles may be proved
congruent by asa or aas.
6. aa JKZ XKZ
c
e asa
B
:
:
:
:
∠ ≅ ∠
7.
8.
9.
parallelogram
congruennt
reflexive property
definition of a
10.
11.
a
D
:
: bbisector
definition of a paralle
12.
13.
B rV tV
e
:
:
≅
llogram
alternate interior angles14.
15.
c
a asa
:
:
Test 26Test 261. e : one congruent angle
is already given
22.
3.
4.
5.
6.
B
D sss
a sas
B aas
:
:
:
:
Pythagorean theorem
BB Ha
c
B
:
:
:
7.
8.
definition of a rectangle
opposite ssides of a rectangle
are congruent (aPt)
re9. c : fflexive property
definition of a mi
10.
11.
a HL
e
:
: ddpoint
PM cPctrc
For
12.
13.
14.
B MQ MQ
D LL
a rM
:
:
: :
≅
≅
#14 and #15, you may
assume the figure inot ss
a rectangle.
is a right angle:
all o
15. B nrQ: ∠
tthers may be proved
congruent with cPctrc.
Test 261. e : one congruent angle
is already given
22.
3.
4.
5.
6.
B
D sss
a sas
B aas
:
:
:
:
Pythagorean theorem
BB Ha
c
B
:
:
:
7.
8.
definition of a rectangle
opposite ssides of a rectangle
are congruent (aPt)
re9. c : fflexive property
definition of a mi
10.
11.
a HL
e
:
: ddpoint
PM cPctrc
For
12.
13.
14.
B MQ MQ
D LL
a rM
:
:
: :
≅
≅
#14 and #15, you may
assume the figure inot ss
a rectangle.
is a right angle:
all o
15. B nrQ: ∠
tthers may be proved
congruent with cPctrc.
Test 27Test 271. c : have the same shape but
not the same ssize
corresponding sides are congruent2.
3.
a
B t
:
: wwo
c
sets of congruent angles
ratio of short 4. : llegs is also 13
similar
they a
5.
6.
7.
D
a
D
:
:
:
810
45
=
rre similar:
36two cong
, ,
:
510
612
12
and all
c
=
8. rruent angles proves
similarity, not congruence
9..
10.
11.
e
e
B
:
:
:
210
15
=
reflexive property
perpendicuular lines form
right angles
12. a XsY and rsQ:
:
:
are
e
a
similar
by aa
vertical angles
alt
13.
14. eernate interior angles
aa postulate15. c :
Test 271. c : have the same shape but
not the same ssize
corresponding sides are congruent2.
3.
a
B t
:
: wwo
c
sets of congruent angles
ratio of short 4. : llegs is also 13
similar
they a
5.
6.
7.
D
a
D
:
:
:
810
45
=
rre similar:
36two cong
, ,
:
510
612
12
and all
c
=
8. rruent angles proves
similarity, not congruence
9..
10.
11.
e
e
B
:
:
:
210
15
=
reflexive property
perpendicuular lines form
right angles
12. a XsY and rsQ:
:
:
are
e
a
similar
by aa
vertical angles
alt
13.
14. eernate interior angles
aa postulate15. c :
Test 28Test 281. c : moving and changing
shapes on a grid
22.
3.
4.
5.
a
e reflection
c
:
:
:
translation
B: dilation
fllipped
rotation
counterclockwise
degr
6.
7.
8.
a
e
a
:
:
: eees
reflection
translation of 5 spaces
9.
10.
1
c
B
:
:
11.
12.
13.
c
a r
:
:
rotation of 90º around
the origin
ee
D
:
:
none
V; each point on figure Q has
been
14.
mmoved to the left 5 and up 2.
Q has bee15. B s: ; nn translated and
rotated, so its transformationn
includes rotation.
test 28 - Unit test i i i
soLUtionsGeoMetrY 237
Test 281. c : moving and changing
shapes on a grid
22.
3.
4.
5.
a
e reflection
c
:
:
:
translation
B: dilation
fllipped
rotation
counterclockwise
degr
6.
7.
8.
a
e
a
:
:
: eees
reflection
translation of 5 spaces
9.
10.
1
c
B
:
:
11.
12.
13.
c
a r
:
:
rotation of 90º around
the origin
ee
D
:
:
none
V; each point on figure Q has
been
14.
mmoved to the left 5 and up 2.
Q has bee15. B s: ; nn translated and
rotated, so its transformationn
includes rotation.
Test 29Test 291.
2.
3.
4.
a
c
D
a
:
:
:
:
triangles
right
cosine
tangeent
none of the above5.
6.
7.
8.
9.
e
B Bc
B aB
e ac
a
:
:
:
:
: 55 310
32
5 310
32
5
5 3
1
3
3
3 3
3
9
33
3
=
=
= = =
=
10.
11.
12.
a
D
c
:
:
: 00
453534
º
:
:
:
13.
14.
15.
c
a
B
Test 291.
2.
3.
4.
a
c
D
a
:
:
:
:
triangles
right
cosine
tangeent
none of the above5.
6.
7.
8.
9.
e
B Bc
B aB
e ac
a
:
:
:
:
: 55 310
32
5 310
32
5
5 3
1
3
3
3 3
3
9
33
3
=
=
= = =
=
10.
11.
12.
a
D
c
:
:
: 00
453534
º
:
:
:
13.
14.
15.
c
a
B
Test 30
θ 4
3
5
Test 301.
2.
3.
4.
B ecant
D ant
B
e
: cos
: sec
:
:
cotangent
noone of the above:
it is the cotangent
secan5. a : tt
cosecant6.
7.
8.
9.
10.
11
c
D ca
B ac
a ca
c
:
:
:
:
: 2 32
31
3= =
..
12.
13.
14.
e
B
D
e
:
:
: sin cos
:
42
2
42
2
2 2 1
4 24
21
=
=
+ =
= =
θ θ
22
44
115. c : =
Test 301.
2.
3.
4.
B ecant
D ant
B
e
: cos
: sec
:
:
cotangent
noone of the above:
it is the cotangent
secan5. a : tt
cosecant6.
7.
8.
9.
10.
11
c
D ca
B ac
a ca
c
:
:
:
:
: 2 32
31
3= =
..
12.
13.
14.
e
B
D
e
:
:
: sin cos
:
42
2
42
2
2 2 1
4 24
21
=
=
+ =
= =
θ θ
22
44
115. c : =
Unit Test IIIUnit Test IIII
1.
2.
3.
axiom or postulate
dilation
reeflection
tangent
secant
similar
sphere
c
4.
5.
6.
7.
8. ootangent
Unit test i i i - Unit test i i i
soLUtions GeoMetrY238
II
III
1.
2.
3.
L H
L H
both
= = ( ) =
= = =
=
4 3 2 4 8
2 3 2 3 2 2 6
;
;
legs 66 2
2
61
6
102
5
5 3
= =
= =
=
4. short leg
long leg
IV Find length of hypotenuse:
42 52 2
16 25 2
41
+ =
+ =
=
H
H
HH
H
2
41
4
41
4 41
41 41
4 4141
5
41
5 41
41
=
= = =
= =
1.
2.
sin
cos
θ
θ441
5 4141
45
414
415
54
=
=
=
=
=
3.
4.
5.
6.
tan
csc
sec
cot
θ
θ
θ
θ
IV Find length of hypotenuse:
42 52 2
16 25 2
41
+ =
+ =
=
H
H
HH
H
2
41
4
41
4 41
41 41
4 4141
5
41
5 41
41
=
= = =
= =
1.
2.
sin
cos
θ
θ441
5 4141
45
414
415
54
=
=
=
=
=
3.
4.
5.
6.
tan
csc
sec
cot
θ
θ
θ
θ
V X X X
X X X
X
+( ) + +( ) + −( ) =
+ + −( ) + + =
− + =
4 2 6 5 12
2 5 4 6 12
2 10 1222 2
1
4 1 4 3
2 6 2 1 6 2 6 4
5
− == −
+ => −( ) + =
+ => −( ) + = − + =
−
XX
X
X
X ==> − −( ) =
+ =+ =
5 1 5
32 42 52
9 16 25
sides are 3, 4 and 5.
225 25= : true
since the Pythagorean theorem
appliees, this is a right triangle.
VI Please note: the proofs given
here may not be the only valid
options. as long as each
statemeent is based on given
information, valid postullates,
definitions and theorems, or on
statemennts made previously
within the proof, the studeent's
proof can be considered correct.
ac
Dca given
Bac Dac
bisects BaD given
Bca
a bi
∠
∠ ≅ ∠
∠ ≅ ∠ ssector divides
the angle into equal
parts
ac ≅ acc reflexive property
Bca is a right angle given∠
∠DDca is a right angle supplementary angles
Bac ≅Dac La
1.
soLUtionsGeoMetrY 239
Unit test i i i - FinaL eXaM
5. BDc
i
is a triangle
hypotenuse
30 60 90
8
º º º− −
= nn
BD in
Bc
m
short leg
long leg
( ) = =
( ) =
∠
8 2 4
4 3
14
÷
6. == − ∠ =
− =
180 5
180 30 150
º
º º º
m
7. no, line ec is not parrallel to line ac
point e
Let X ength of ae
8.
9. = l
2208 4
8 4 20
8 80
10
=
= ( ) ( )=
=
X
X
X
X
10. First find length oof ac
eac is a triangle,
so the
:
º º º 30 60 90− −
llong leg is 3 times
the short leg or 10 3
aB ac= − BBc = − =10 3 4 3 6 3 III
1. ce ca
aBc cDe
Dce
≅
∠ ≅ ∠
∠ ≅ ∠
given
given
acB vertical angless
aBc ≅ cDe aas
2.
aB Bc≅
∠
∠
given
Bec is a right angle given
Bea is a riight angle supplementary
Be reflexive property≅ Be
aBe ≅
≅
cBe HL
ae ce cPctrc
IV.
V V r
cm
= ( ) ( )
=
43
3 43
3 14 33
113 04 3
π ≈ .
.
if the fractionaal value of is used,
the answer would be 113.
π
114 cm3.
2.
BD ce
ace
given
aBD corresponding ∠ ≅ ∠
aangles
B cae∠ ≅ ∠aD reflexive
property
ace ~ aBD a aa
3.
m aDc m BcD
Dc
∠ = ∠ =
≅
90º definition of a
rectangle
Dc reeflexive property
aD opposite sides of a
rect
≅ Bc
aangle are
congruent (aPt)
aDc sas or LL ≅ BcD
Final ExamFinal ExamI
1.
2.
3.
4.
cosine
obtuse
arc
complementary
55.
6.
7.
8.
9.
plane
trapezoid
cube
collinear
congruent
110. perimeter
II1.
2.
3.
trapezoid
12
corresponding an
∠
∠ = ∠ =m m6 8 60º
ggles
4. m m m∠ = − ∠ + ∠( ) =
− +( ) =
−
5 180 4 6
180 60 90
180 1
º
º º º
º 550 30º º= 5. BDc
i
is a triangle
hypotenuse
30 60 90
8
º º º− −
= nn
BD in
Bc
m
short leg
long leg
( ) = =
( ) =
∠
8 2 4
4 3
14
÷
6. == − ∠ =
− =
180 5
180 30 150
º
º º º
m
7. no, line ec is not parrallel to line ac
point e
Let X ength of ae
8.
9. = l
2208 4
8 4 20
8 80
10
=
= ( ) ( )=
=
X
X
X
X
10. First find length oof ac
eac is a triangle,
so the
:
º º º 30 60 90− −
llong leg is 3 times
the short leg or 10 3
aB ac= − BBc = − =10 3 4 3 6 3
soLUtions GeoMetrY240
VI
VI
sa
cm
= ( ) ( ) + ( )( ) + ( ) ( ) =
+ + =
2 2 5 2 2 7 2 5 7
20 28 70 118 2
II 360
360 45 8
º
º º
total of all angles
sides; oct÷ = aagon
VIII1. 3 2 4 22 3 4 2 22
12 44 12 4 11 12 2
( )( ) = ( ) ( ) =
= = ( ) 111
24 11
4
3
2 6
2
4 3
3 3
2 31
4 3
9
2 31
4 33
2 31
4 33
2
=
− = − =
− = − =
−
2.
33 31 3
4 33
6 33
4 3 6 33
2 33
3 5 5 3 1 5 2
( )( ) = − =
− = −
− + = − +( ) = −3. 55
2 3 4 1
2 3 2 1 2 3 3
88
8
4. + + + =
+ + + = + +
= => =
=
=
IX c d dd
π π πππ
ππ
dd
radius
c
= =( )12
8 4
X heck with ruler:
smallerr segments should
each measure 2 inches.
he mXI t eeasure of a central angle
is equal to the measuure of the
arc it intercepts.
the meas
m aXc∠ = 98º
uure of an inscribed
angle is half the measure oof the
arc it intercepts.
m aBc∠ = =98 2 49º º÷
XII L
L
L
L
2 22 52
2 4 252 21
21
+ =
+ =
=
=
XIII start by drawing a diagram.
sine is 35
= oppossitehypotenuse
so we know that the hypotenuse
iis 5, and one leg is 3.
L2 32 52
2 9 252 16
4
+ =
+ =
==
L
LL
sin csc
cos sec
so other leg is 4
θ θ
θ θ
= =
= =
35
53
45
54
ttan cotθ θ= =34
43
FinaL eXaM
4
3
5θ
