Circle - Holy Cross School resources/Mathematics Conten… · Line through a point on a circle is a...
Transcript of Circle - Holy Cross School resources/Mathematics Conten… · Line through a point on a circle is a...
Grade 11 CAPS
Mathematics
Video Series
Circle
Geometry II
In this Video we will :
Lessons linked to this Video
and theorems that are
linked to
and related .
Investigate prove
cyclic quadrilaterals
solve ride Lessrs on 1
and theorems that are
linked to and
related .
Investigate prove
tangents of circles
solve riders Lesson 2
Grade11 CAPS
Mathematics
Video Series
Lesson 1
Theorems
On
Cyclic
Quadrilaterals
In this lesson we will :
Outcomes for Lesson 1
Recap the linked quadrilaterals and cyclic quadrilaterals. terminology
Solve riders related to Theorems 1, 2 and its converses.
Prove that :
,
(
If the sum of the two opposite angles of a
quadrilateral is equal to 180 its vertices are concyclic.
Converse o f Theorem 1)
Investigate and prove that :
(
The exterior angle of a cyclic
quadrilateral is equal to opposite interior T ang heorle. em 2)
Prove that :
(
If the exterior angle of a quadrilateral is equal
to its to opposite interior angle, the quadrilateral is cyclic.
Converse of Theorem 2)
Investigate and prove that :
(
The sum of the opposite angles
of a cyclic quadrilateral is equal t Theoro 180 . em 1)
Any four sided polygon is a .quadrilateral
Quadrilateral Terminology (Undefined terms)
1, 2, 3 and 4 are the
of quadrilateral .
1 2 3 4 360
ABCD
interior angles
Know that :
5, 6, 7 and 8 are the exterior angles
of quadrilateral .
5 6 7 8 360
ABCD
Know that :
Points lying on a circle are .
A quadrilateral having its vertices on a circle
is called a .
concyclic
cyclic quadrilateral
Inscribed angles on the same side of a chord are equal.
Revisit Theorem 5 (Video Lesson : Circle Geometry 1)
Quadrilateral with , , and are concyclic or
Quadrilateral with is a cyclic quadrilateral or
If
ABCD ADB ACB A B C D
ABCD ADB ACB ABCD
Consider version 4 of Theorem 5 in formulation of converse :
the line segment joining two points subtends equal angles at two other points
on the same side of it, the four points must be concyclic.
:Converse of Theorem 5 (Proved in next slide)
1 Inscribed angles on the same side of a chord are equal (original).
2 Inscibed angles on a chord in the same segment are equal.
3 , , and concyclic .
4 a
A B C D ADB ACB
ABCD
Theorem 5 can be reformulated :
cyclic quadrilateral .ADB ACB
If line segment joining two points subtends equal angles
at two other points on the same side of it, the four points must be concyclic.
Converse Theorem :
s subtending equal and at
and on the same side of .
, , , and are a concyclic.
Draw and assume that .
BC BDC BAC
D A BC
A B C D
DBC A DBC
Given :
Aim is to prove that :
Construction :
Let cut at and draw . DBC BA M MC
:
1 inscribed
Assume that and
s n
1
o
M DBC A DBC
BDC BC
Proof
A quadrilateral has is a cyclic quadrilateralABCD BAC BDC ABCD
Combining Theorem 5 and its Converse we have :
Ext. 1 of int. opp. AMC A
This is a contradiction because 1 of 1AMC A ACM A
Assumption 1 is incorrect and thus , , and are concyclic.A B C D
Sum of opposite angles of a cyclic
quadrilateral is equal to 180 (or supplementary).
Investigation :
for more
such investigations!
GeoGebra
Suggested Conclusion :
180Cyclic quadrilateral
180
A CABCD
B D
Conjecture 1 :
Any with cyclic quadrilateral .
180
180
Draw and
O ABCD
A C
B D
DO BO
Given :
Aim is to prove that :
Construction :
:
1 2 central 2 inscribed C
Proof
s2 2 2 1 360 round a pointA C
sBut 360 of a quadrilateral A C B D
Sum of opposite angles of a cyclic
quadrilateral is equal to 180 (or supplementary).
Theorem :
180Cyclic quadrilateral
180
A CABCD
B D
Theorem 1 :
2 2 central 2 inscribed A
180A C
180 180 B D A C
Assumption in 1 was incorrect must fall on passes throug M D ABC D
If the sum of the two opposite angles of a
quadrilateral is equal to 180 its vertices are concyclic.
Theorem :
but 180 givenD B
1 which is impossible 1 1D D MCD D
Any quadrilateral having 180 .
is a cyclic quadrilateral.
Draw and assume that .
Let cut at and dra
ABCD B D
ABCD
ABC D ABC
ABC AD M
Given :
Aim is to prove that :
Construction :
w . MC
Assume : that and 1M ABC D ABC Proof
1 180 , , and are concyclicB A B C M
Quadrilateral with opposite angles supplementary Quadrilateral is cyclic
Converse of Theorem 1 :
Quadrilateral is cyclic sum of opposite angles of quadrilateral is 180
Combining Theorem 1 and its Converse we have :
The exterior angle of a cyclic quadrilateral
is equal to the interior opposite angle.
Investigation :
From investigations we make the :
Exterior angle of a cyclic quadrilateral
be equal to opposite interior angle.
conjecture
seems to
for more
such investigations.
GeoGebra
Cyclic quadrilateral and
with .
Exterior
interior opposite
ABCD
AE B AE
CBE
CDA
Given :
Aim is to prove that :
2 180 straight CBE ABE
Quadrilateral cyclic exterior interior opposite
Theorem 2 :
The exterior angle of a cyclic quadrilateral
is equal to the interior opposite angle.
Theorem :
s
:
1 2 180 sum of int. opp. of cyclic quad.
Proof
1 2 2 1CBE CDA CBE
Quadrilateral having drawn
with and .
is a cyclic quadrilateral.
ABCD DC
E DC BCE DAB
ABCD
Given :
Aim is to prove that :
:
1 180 straight BCE DCE
Proof
If the exterior angle of a quadrilateral is equal
to its interior opposite angle, the quadrilateral is cyclic.
Theorem :
1 180 given:
or 2 1 180
DAB BCE DAB
sQuadrilateral is cyclic sum of opp. 180ABCD
A quadrilateral is cyclic Exterior angle is equal to opposite interior angle
Combining Theorem 2 and its Converse we have :
exterior interior opposite Quadrilateral cyclic
Converse of Theorem 2 :
s
1
on chord
or similar results linked to chords , and
BAC BDC BC
AB AD DC
ABCD is a cyclic quadrilateral if we can prove that :
Sufficient conditions for a quadrilateral to be cyclic
2
180 Opposite angles supplementary
or 180
A C
B D
ABCD is a cyclic quadrilateral if we can prove that :
3
Ext. Opp. int.
or similar results linked to , , and
their respective opp. ext. angles.
DCE A
B C D
ABCD is a cyclic quadrilateral if we can prove that :
Riders mainly linked to Theorem 1 and its converse
Quadrilateral is cyclic sum of opposite angles of quadrilateral is 180
Theorem 1 and its Converse :
s
180 60 70 50
Sum of in
e
s
180 180 65 115
Opp. of cyclic quad.
b a
s
5 3 180
Opp. of cyclic quad.
c d
36
and 60
c
d
s
70 180 110
opp. of cyclic quad.
f f
s
60 35 180
opp. of cyclic quad.
g e
180 60 50 35
35
g
13065
2
1 at circum at centre
2
a
Find the value of each letter in the riders.
Rider 1 Rider 2Rider 3
A quadrilateral is cyclic Exterior angle is equal to opposite interior angle
Theorem 2 and its Converse :
Riders linked to mainly Theorem 2 and its converse
113
ext. opp. int.
f
s
180 95
co-int.
b a
88
ext. opp. int.
d
s
180 67 113
opp. of cyclic quad.
e
85
ext. opp. int.
a
Find the value of each letter in the riders.
s
180 85
opp. of cyclic quad.
c b
92
ext. opp. int.
g
Rider 1
Rider 2 Rider 3
Tutorial 1: Find the value of the letters in each of the following riders
PAUSE Video
• Do Tutorial 1
• Then View
Solutions
Tutorial 1: Riders 1 and 2: Suggested Solutions
Determine the values of the angles represented by the letters.
s6 180 Opp. of cyclic quad.
30
a
a
2 at centre 2 at circum
60
b a
b
s4 180 Opp. of cyclic quad.
45
c
c
s4 80 180 Opp. of cyclic quad.
4 100
25
d
d
d
Tutorial 1: Riders 3 and 4: Suggested Solutions
Determine the values of the angles represented by the letters.
s 1 180 70 Opp. of cyclic quad.
1 110
s 2 180 110 70 sum of in e
sBut 2 35 base of isosceles e
1 ext. opp. int. f
sBut 1 85 180 co-interior
1 180 85 95
1 95f
Tutorial 1: Rider 5: Suggested Solution
Determine the values of the angles represented by the letters.
88 ext. opp. int. g
s180 co-interior
180 180 88 92
g h
h g
s180 Opp. of cyclic quad.
180 92 88
i h
i
iCan you see an alternative method to find ?
Tutorial 1: Rider 6: Suggested Solution
Determine the values of the angles represented by the letters.
s 1 2 180 co-interior
180 1 2 180 72 90 18
k
k
s
2 90 18 108
3 180 2 Opp. of cyclic quad.
k
j k
1 72 ext. opp. int.
2 90 in semi-circle
s 3 alternate k
Label angles
180 108 18 3 18 54j k j
Grade11 CAPS
Mathematics
Video Series
Lesson 2
Theorems
On
Tangents
In this lesson we will :
Outcomes for Lesson 2
Recap the and linked to tangents. terminology axioms
Investigate and prove ( :
Line through a point on a circle is a tangent to the circle line is perpendicular to the radius.
Theorem 3 and its converse)
⇔
Solve riders related to Theorems 3, 4 and 5 and its converses.
Prove the following linked to theorem 3 and its converse :
1 Tangents drawn from an external point to a circle are equal.
2 The bisector of the angle between the two tangents passes
corollaries
through the centre of the circle.
3 The line segment joining the centres of two circles cutting each other is the perpendicular
bisector of the common chord.
Investigate and prove ( :
The bisectors of the interior angles of a triangle are concurrent.
Theorem 4)
Investigate and discuss circle and circles of a given triangle. inscribed escribed
Investigate and prove ( :
A line is a tangent to a circle
Angle formed between a line, that is drawn through the end point of a chord and the chord,
is equal to the an
Theorem 5 and its converse)
gle subtended by the chord in the alternate segment.
Terminology linked to tangents
is a as it cuts in two points and . AB O A Bsecant line
If approaches until it may be considered
to coincide with , the secant is a to
the at the point of coincidence.
B A
A tangent
The point of coincidence is called the
.point of contact
We may say:
A tangent is a line (line segment) which
has only one point in common with the .
Shortest distance from a point to a line
Axiom:
1 The is the shortest
line segment from a point to a line.
2 Conversely, the shortest line segment
from a point to a line is the perpendicular.
perpendicular
From investigations we make the :
If a line through a point on a is a tangent to the
then the line is perpendicular to the radius.
conjecture
for more
such investigations.
GeoGebra
If a line through a point on a is a tangent
to the , then the line is perpendicular to the radius.
Investigation :
If line through a point on a is a tangent to the
then the line is perpendicular to the radius.
Theorem 3 :
If a line through a point on a is a tangent
to the then the line is perpendicular to the radius.
Theorem :
is a tangent with the point of contact.
.
Draw with
AB O P
AB OP
OT T AB
Given :
Aim is to prove that :
Construction :
:
lies outside is contact pointT O P
Proof
is the radius of OP OT OP O
is the shortest of all segments drawn from to .OP O AB
Perpendicular is shortest distance between a point and lineOP AB
A line drawn through any point of a
circle perpendicular to the radius of the is a tangent to the .
Converse of Theorem 3 :
A line drawn through any point of a circle
perpendicular to the radius of the is a tangent to the .
Theorem :
and any point on this circle.
with .
is a tangent to at .
Draw with
O P
P AB OP AB
AB O P
OT T AB
Given :
Aim is to prove that :
Construction :
: given
is shortest distance from to
AB OP
OT OP OP O AB
Proof
radius of lies outside OT O T O
Any point on , except , lies outside is a tangent to AB P O AB O
Line through a point on a circle is a tangent to the circle line is perpendicular to the radius.
Combining Theorem 3 and its converse :
⇔
linked to Theorem 3 and its converse.Corollories
Tangents drawn from an external point to a circle are equal.Corollory 1:
The bisector of the angle between the two tangents
passes through the centre of the circle.
Corollory 2:
The line segment joining the centres of two circles cutting each
other is the perpendicular bisector of the common chord.
Corollary 3 :
and tangents to AP BP O
AP BP
bisectsPC APB
O PC
and with common chord
and
O P AB
OP AB AC BC
Proof of Corollary 1.
Tangents drawn from an external point to a circle are equal.Corollory 1:
and tangents to AP BP O
AP BP
Given :
Aim is to prove that :
sIn and ,OAP OBPProof :
radii
common
tangent radius
AO BO
OP OP
OAP OBP
90 , ,AOP BOP s s AP BP
and tangents to AP BP O AP BP
Proof of Corollary 2.
The bisector of the angle between the two tangents
passes through the centre of the circle.
Corollory 2:
bisectsPC APB
O PC
Given:
Aim is to prove that:
Thus we assume that
is not a
Assume that
bisector of .
PO
APB
O PCProof :
radii
common
tangent radius
AO BO
OAP OBP OP OP
OAP OBP
APO BPO
and tangents to and bisect AP BP O PC APB O PC
also bisectsPO APB O PC
Proof of Corollary 3.
The line segment joining the centres of two circles cutting each
other is the perpendicular bisector of the common chord.
Corollary 3 :
and with common chord
and
O P AB
OP AB AC BC
Given:
Aim to prove that:
sIn and ,ACP BCPProof :
proved
common
proved
AP BP
PC PC
APC BPC
, ,ACP BCP s s
90ACP BCP
AC BC
and with common chord
and
O P AB
AB OP AC BC
From investigations we make the :
The bisectors of the three interior angles of any triangle
are concurrent ( or intersect in a common point).
conjecture
for more such investigations.GeoGebra
The bisectors of the interior angles of a triangle are concurrent.
Investigation :
The bisectors of the interior angles of a triangle are concurrent.Theorem 4 :
The bisectors of the interior angles of
a triangle are concurrent.
Theorem :
Any .
Bisectors of the angles are concurrent.
Draw and the bisectors of and meeting in .
Draw and perpendiculars , and to , and
ABC
BO AO B A O
OC OD OE OF AB AC
Given :
Aim is to prove that :
Construction :
respectively.BC
:
, ,BDO BFO s OD OF
Proof
, ,ADO AEO s OD OE
OF OD OE
In and :CEO CFO
90 construction
common
proved
CEO CFO
OC OC
OE OF
, ,90CEO CFO s s
is a angle bisector of meeting other two angle bisectors in . ECO FCO CO C O
Bisect any two angles and
determine the in-centre .O
circle of a gi .ven triangleConstruction : Inscribed
Drop a perpendicular to any of
the three sides to determine the
radius of the inscribed .r OD
With as centre and
as radius draw inscribed .
O OD , and are tangents
AB AC BC
r OE OD OF
Bisect any two of indicated angles to determine the centre.
Drop perpendicular to determine the radius .
Draw escribed circle (Take note of three tangents).
r OF
circles of a given triangle.Construction : Escribed
Sketches of other two left as an exercise.
Take note of tangents in each case.
Three possibilities.
Example 1: Lengths of tangents and inscribed circles.
The inscribed touches , and of at , and respectively.O AB BC AC ABC D F E
1 If 8 units, 12 units and 9 units
determine the lengths of , and .
AB BC AC
AD BF CF
, and
Tangents from exterior points are equal
AD AE CE CF BD BF
8 8 1
9 9 2
and 12 3
AD BD AD BF
AE EC AD CF
CF BF
2 1 1 4
3 4 2 13 6.5
CF BF
CF CF
Back substitution
12 12 6.5 5.5
8 8 5.5 2.5
BF CF
AD BF
Check correctness :
2.5 5.5 8
2.5 6.5 9
5.5 6.5 12
AB AD DB
AC AE EC
BC BF FC
Example 2 : Lengths of tangents and inscribed circles.
The inscribed touches , and of at , and respectively.
Assume that , and .
O AB BC AC ABC D F E
AB c AC b BC a
2 If prove that .2
a b cS AD S a
, and
Tangents from exterior points are equal
AD AE CE CF BD BF
Perimeter of Perimeter of 2
2 2
ABC a b cS ABC S
2
2 2 2 2 ; ;
S AD AE BD BF CE CF
S AD BF CF AD AE BD BF CE CF
S AD BF CF
S AD BF CF AD BF CF AD a AD S a
Example 3 : Lengths of tangents and inscribed circles.
The inscribed touches , and of at , and respectively.
Assume that , and .
O AB BC AC ABC D F E
AB c AC b BC a
area of
areas of
ABC
BOC AOC AOB
3 If prove that area of .2
a b cS ABC rS
2 2 2
ar br cr
2
a b cr rS
Tutorial 2: Tangents, Inscribed and Escribed circles
1 The sides of a quadrilateral are tangents to .
Prove that .
ABCD O
AB CD AD BC
2 is the centre of the inscribed circle of .
The circle touches , and at , and
respectively. , and are drawn.
2.1 Make a rough sketch.
2.2 Prove that
O ABC
AB BC CA P Q R
OP OQ OR
.2
B CPQR
3 is an escribed circle of with radius .
If 2 , prove that:
3.1 .
3.2 and
3.3 area of .
DEF ABC r
S a b c
AD AF S
CE S b
ABC r S a
PAUSE Video
• Do Tutorial 2
• Then View Solutions
Tutorial 2: Problem 1: Suggested Solution
1 The sides of a quadrilateral are tangents to .
Prove that .
ABCD O
AB CD AD BC
AB CD
AE EB DG CG
from known results - see figureAH BF HD CF
commutative & associative
properties for additionAH HD BF CF
AD BC
: Proof
2 2
PQR PQO OQR
B C
Tutorial 2: Problem 2: Suggested Solution
2 is the centre of the inscribed circle of .
The circle touches , and at , and
respectively. , and are drawn.
2.1 Make a rough sketch.
2.2 Prove that
O ABC
AB BC CA P Q R
OP OQ OR
.2
B CPQR
s180 Opp. of cyclic quad.POQ B
s180 Sum of in PQO QPO POQ OPQ
2 180 180 2
BPQO B OP OQ PQO
Similarily because is a cyclic quad. 2
CROQC OQR
sis a cyclic quadrilateral Opp. supplementaryPOQB
Tutorial 2: Problem 3.1: Suggested Solution
3.1 is an escribed circle of with radius .
If 2 , prove that .
DEF ABC r
S a b c AD AF S
, and
Tangents from exterior points are equal
AD AF BD BE CE CF
2 2AD AF AD AF AD AF
2 2AD AF AB BD AC CF
2 2 ;AD AF AB BE AC EC BD BE CF EC
2 2AD AF AB AC BE EC
AB AC BC c b a
2
a b cAD AF S
Tutorial 2: Problem 3.2: Suggested Solution
3.2 is an escribed circle of with radius .
If 2 , prove that .
DEF ABC r
S a b c CE S b
AF AC CF
AC EC CF EC
EC AF AC
Proved in 3.1 that .AF S
EC AF AC
S b AC b
: Proof
Tutorial 2: Problem 3.3: Suggested Solution
3.3 is an escribed circle of with radius .
If 2 , prove that area of .
DEF ABC r
S a b c ABC r S a
area area and
area area and
area area and
BDO BEO BD BE h r
CEO CFO EC CF h r
DAO FAO AD AF h r
area
2 area 2 area 2 area
ABC
ADO BDO CEO
AD r BD r CF r
r AD BD CF
r AB BD BD CF
r AB CE
2
2 1
S a b c
b c S a
...
1 : 2r S a S r S a From ( ; )r c S b CE CF AF S
From investigations we make the :
If a line is a tangent to a circle then the angle between a chord
and the tangent, drawn at the point of contact of the chord, is equal
to the inscribed angle
conjecture
which the chord subtends in the alternate segment.
for more
such investigations.
GeoGebra
The angle between a tangent and a chord, drawn
at the point of contact of the chord, is equal to the inscribed
angle which the chord subtends in the alternative segment.
Investigate :
Angle between tangent and chord is equal to the inscribed angles
which the chord subtends in alternate segment.
Theorem 5 :
Angle between tangent and chord is equal to the inscribed
angles which the chord subtends in alternate segment.
Theorem :
Any with chord and
tangent with point of contact .
and
.
Draw diameter and .
O BD
AC B
DBC F
ABD M
BE DE
Given :
Aim is to prove that :
Construction :
:
90 angle in a semicircle
1 90 sum s in a
EDB
E
Proof
180 180ABD DBC ABC
1 90
tangent and radius
DBC
BC OB
inscribed s on chord DBC E F BD
180 is a cyclic quad.F M FBMD
180
180
ABD DBC
F DBC F
M
If the angle formed between a line, that is drawn
through the endpoint of a chord, and the chord, is equal to the angle subtended
by the chord in the alternate segment, then the line
Converse Theorem :
is a tangent to the circle.
and chord such that
in alternate arc.
is a tangent to at .
Draw a tangent at .
ABC AB
TAB ACB
AT ABC A
AE A
Given :
Aim is to prove that :
Construction :
:
angle between tangent and chord
inscribed in alternate arc
AE ABACB EAB
Proof
But GivenACB TAB
EAB TAB
This is only possible if and coincide.AE AT
is a tangent to at .AT ABC A
If the angle formed between a line, that is drawn
through the endpoint of a chord, and the chord, is equal to the angle subtended
by the chord in the alternate segment, then the
Converse of Theorem 5 :
line is a tangent to the circle.
Sufficient condition for a line (segment) to a circle to be a tangent.
1 If
then is a tangent to .
BA OA
BA O
2 If or
then is a tangent to .
CBD A ABE C
ED ABC
Riders linked to Tangents
9018
5a
90
angle in semi-circle
d
40
angle between tangent & chord
equal to angle in opp. segment
b
80
isosceles
g f
180
sum 180
180 160 20
e g f
e
5 90
radius tangent
sum 180
a
Find the value of the angles indicated by each letter in the riders.
Where applicable is the centre of the circle and a tangent.O AB
90 40 50
radius tangent or sum 180
c
80
angle between tangent & chord
equal to angle in opp. segment
f
More Riders linked to Tangents
70
1 at circum. at centre
2
q
180 2 48 84
angles in isosceles
j
30
angles on same chord
l
90 60
angle in semi-circle
and sum 180
m l
48
tangent & chord
in opp. segment
i
Find the value of the angles indicated by each letter in the riders.
Where applicable is the centre of the circle and a tangent.O AB
84
tangent & chord
in opp. segment
h j
30
chord & tangent
in opp. segment
k
70
chord & tangent
in opp. segment
n q
90 20
alt. s and radius tangent
p q
Tutorial 3: Riders linked to Tangents
PAUSE Video
• Do Tutorial 3
• Then View Solutions
Find the value of the angles indicated by each letter in the riders.
Where applicable is the centre of the circle and a tangent.O AB
Tutorial 3: Rider 1: Suggested Solution
Find the value of the angles indicated by each letter in the rider.
Where applicable is the centre of the circle and a tangent.O AB
65
tangent & chord in opp. segment
a
55
tangent & chord in opp. segment
b
Tutorial 3: Rider 2: Suggested Solution
Find the value of the angles indicated by each letter in the rider.
Where applicable is the centre of the circle and a tangent.O AB
38
tangent & chord in opp. segment
Note: and angles on same chord
d e
d e
90 38 52
radius tangent
c
Tutorial 3: Rider 3: Suggested Solution
Find the value of the angles indicated by each letter in the rider.
Where applicable is the centre of the circle and a tangent.O AB
1
2
1 4
tangent & chord in opp. segment
f
2 alt. 1 4
parallel line segments
f
2 10 4 180
straight angle
f f 4 10 4 180
18 180
10
f f f
f
f
Tutorial 3: Rider 4: Suggested Solution
Find the value of the angles indicated by each letter in the rider.
Where applicable is the centre of the circle and a tangent.O AB
1
50
tangent & chord in opp. segment
h
1 2 100
at centre 2 at circum.
h
180 1
2
isosceles
sum 180
g
180 10040
2g
Tutorial 3: Rider 5: Suggested Solution
Find the value of the angles indicated by each letter in the rider.
Where applicable is the centre of the circle and a tangent.O AB
40
tangent & chord in opp. segment
k
40 60 100
tangent & chord in opp. segment
j
180
opp. s in cyclic quad. supplementary
i j
180 100 80i
End of the Second Video on Circle Geometry
REMEMBER!
•Consult text-books for additional examples.
•Attempt as many as possible other similar examples
on your own.
•Compare your methods with those that were
discussed in this Video.
•Repeat this procedure until you are confident.
•Do not forget:
Practice makes perfect!