Geometric Similarities

Math 416

Geometric Similarities Time Frame

1) Similarity Correspondence 2) Proportionality (SSS) (Side-side-side) 3) Proportionality (SAS) (side-angle-side) 4) Similarity Postulates 5) Deductions 6) Dimensions 7) Three Dimensions

Similarity Correspondence

Similarity – Two shapes are said to be similar if they have the same angles and their sides are proportional

Note – we see shape by angles & we see size with side length

Consider

A

C

D

B ZY

WX

Similar & Why?8

32

24 16 5

20

15 10

100

80 85

95

80

100 95

85

Proportionality (SSS)

We say the two shapes are similar because their angles are the same and their sides are proportional

We can note corresponding points A X D W B Y C Z

Angles

We note corresponding angles < ADC = < XWZ (95°) < DCB = < WZY (85°) < CBA = < ZYX (80°) < BAD = < YXW (100°)

Notes Hence we would say ADCB XWZY Hence we note corresponding angles < ADC = < XWZ < DCB = < WZY < CBA = < ZYX < BAD = < YXW

Proportionality

Next is proportionality which we will state as a fraction

AD=8 DC=16 CB=32 BA=24 XW 5 WZ 10 ZY 20 YX 15 What is the proportion (not in a

fraction)? 8/5 which is reduced to 1.6

Question #1

Identify the similar figures and state the similarity relationship, side proportion and angle equality

A

C

Z

T

C

B

BIG

SMALL

BIG

MEDMED

SMALL

Notes for Solution By observing you need to establish

the relationship. Look at angles or side lengths Important: An important trick

when comparing angles and sides is that the biggest angles is always across the biggest side, the smallest from the smallest and medium from the medium.

Solution #1

Triangle ABC ˜ TCZ AB = BC = CA TC CZ ZT < ABC = < TCZ < BCA = <CZT < CAB = <ZTC

Important Note

Make sure the middle angle letters are all different because the middle letter is the actual angle that you are looking at.

AC = CA < ACB = < BCA Both the above are the same

Question #2

K R

LT

XQ

MED

MED

SMALL

SMALL

MED

MED

With isosceles (or equilateral triangles) you may get two (or three) different answers). However, you are only required to provide

one.

Solution a for #2 The question is to identify similar figures

and state the similarity relationship, side proportion and graph equality.

QK = KT = QT RX XL RL < QKT = < RXL < KTQ = < XLR < TQK = < LRX

QKT ˜ RXL

Solution b for #2

You can also have another solution Triangle QKT is still congruent to RLX QK = KT = QT RL LX RX < QKT = < RLX < KTQ = < LXR < TQK = XRL

More Notes There are other ways of establishing

similarity in triangles At this point we will abandon reality for

simple effective but not accurate drawings of triangles… (it is not to scale).

Please complete #1 a – o For Question #3, again, state similarity

relationship, side proportion and angle equality.

Question #3

T

27

21

30

453518

Q

RCB

A

If the three sides are proportional to the corresponding three sides in the other triangle, the two will be similar.

Solution Notes

You need to check…

SMALL with SMALL

MEDIUM with MEDIUM

BIG with BIG

Solution #3

ABC ˜ QRT

MedSmall

Big

SmallMed

Big

18 = 21 = 27

30 35 45

0.6 = 0.6 = 0.6; YES SIMILAR

Proportionality SAS

We can also show similarity in triangles if we can find two set corresponding sides proportional and the contained angles equal; we can determine similarity

14°18

A

BC

ZY

X

4214°

15

35

Question #4 Show if the triangle is similar Solution… since <BAC = <XYZ = 14° 18 = 42 15 35= 6/5 6/5 BAC ˜ XYZ Notice BAC = Small, Angle, Big &

compared to Small, Angle, Big

Triangle Similarity Postulates There are three main postulates we use to

state similarity SSS all corresponding sides proportional SAS two sets of corresponding sides and

the contained angle are equal AA two angles (the third is automatically

equal since in a triangle, the interior angle must add up to 180°) are equal

Example #1

Why are the following statements true?

QPT ˜ ZXA

42°84°

84° 54°54°

AA

P T ZA

XQ

Example #2

Why are the following statements true?

KTR ˜ PMN

51°

51°

Solution: since 24/16 = 27/18

R

KT PN

M

27 1618

SAS

24

Example #3

16 12

CB

A

9

24 6

32 PT

K

Since 16 = 24 = 32

6 9 12

8/3 = 8/3 = 8/3S S S

Parallel Lines Facts: If two tranversals intersect

three parallel lines, the segments between the lines are proportional

a

b d

Therefore, a = c

b d

c

Notes

Also note that…

A

C

B BC = 1

AC 2•

•

•

Parallel and the Triangle If a parallel line to a side of a triangle

intersects the other two sides it creates two similar triangles

A

EB

CD

Therefore, ABE ˜

ACD

Question #1

3

4 x

9

Find x

3 = 9

4 x

3x = 36

x = 12

Question #2

C

x

B

40

150

50E

A

D

We note BE // CD

Thus, ABE ˜ ACD

AB = BE = AE

AC CD AD

x = 50 = AE

x+40 150 AD

Question #2 Sol’n Con’t

We only need x = 50 x+40 150 150x = 50(x+40) 150x = 50x + 2000 100x = 2000 x = 20

Proportion Ratio

Consider

1

10

1

10

Dimensions

SIDE SMALL BIG RATIO

SIDE 1D 1 10 1:10 or 1/10

AREA 2D 1 100 1:100 OR 1/100

Dimensions In general in 1D if a:b then in 2D a2 : b2

Ex. In 1D if 5:3 then in 2D? In 2D then 25:9 You can go backwards by using

square root Ex. In 2D if 36:49 then in 1D 6:7

3D or Volume

Consider

1

11 5

5

5

3D or Volume

Small Big Ratio

Side 1D 1 5 1:5 or 1/5

Volume 3D 1 125 1:125 or 1/125

3D or Volume In general in 1D if a:b then in 3D… Then in 3D a3 : b3

Ex. In 1D if 6:7 then in 3D In 3d 216:343 You can go backwards by using the

cube root Ex. In 3D if 27:8 then in 1D In 1D 3:2

Practice

Complete the following

1 D Length 2 D Area 3 D Volume

2:9 4:81 8:729

2:11 4:121 8:1331

5:3 25:9 125:27

3D Question #1

Two spheres have a volume ratio of 64:125. If the radius of the large one is 11cm, what is the radius of the small one?

Big Small r 11 x

3D Ratio 64:125

1D Ratio 4:5

Thus 4 = x

5 115x = 44

x = 8.8

3D Question #2

V=200m3

V = ?

A Base = 100m2A base = 16 m2

Question #2 Solution

Big SmallArea of Base 100 16Volume 200 x

1 Ratio 10 / 4

2 Ratio 100/16

3 Ratio 1000/64Thus 200 = 1000

x 64

1000x = 12800

x = 12.8