Chapter 8 Similarity

232 GeometryBEHS

Mrs. Prescott

Ratios• Ratio – a comparison of 2 numbers in the

same unit of measure• Example: 2 females to 3 males– 2 to 3– 2:3–

2

3

2

Simplifying Ratios• Reduce using common factors as you would

simplify any fraction• Pages 461-462

12.

16.

22.

26.

26

11

52

22

feet

feet

5

8

20

32

20

2

oz

oz

oz

lb

2

1

2

1

2

12

ft

ft

feet

in

2

1

24

12

2

12

in

in

feet

inor

17

7

CF

BD

Proportions

• Proportion – a statement equating 2 ratios• Example:

• Also written - –3:6 = 5:10

10

5

6

3

extremes

means

Proportions• The product of the means is equal to the

product of the extremes–3:6 = 5:10

• The cross products in a proportion are always equal which is why we cross multiply to solve a proportion

10

5

6

3

310 = 65 30 = 30

31065

Solve the Proportion

6

xx

4

3

5

)3(45 xx

1245 xx12x

-4x-4x

Properties of Proportions

1. Cross Product Property – The Product of the extremes equals the

product of the means.–

• Reciprocal Property– If two ratios are equal, then their

reciprocals are also equal. –

., bcadthend

c

b

aIf

.,c

d

a

bthen

d

c

b

aIf

Page 463#56.

7

5

XV

WX

7

5

2

2

k

k

)2(527 kk

kk 10147 -7k -7k

k314 3 3

k3

14324kor

Extended Ratios• Simplify the extended ratio of the 4 sides of

the quadrilateral.• 20:16:40:36 =• 5:4:10:9 20cm

16cm

36cm

40cm

Examples:1. The perimeter of a parallelogram is 42ft, and ratio

of 2 of its unequal sides is 3:4. Find each side length.

4 sides – 3:4: 3:4 or 3x:4x:3x:4x 3x + 4x + 3x + 4x = 42 14x = 42 x = 3

2. If the extended ratio of the angles of a triangle are 5:6:7, find each angle measure.

5x:6x:7x 5x + 6x + 7x = 180

18x = 180 x = 10

Side lengths are 9 ft, 12 ft, 9 ft,

and 12 ft.

Angle measures are 50º, 60º, 70º,

Section 8.2 Problem Solving in Geometry

with Proportions

Page 465232 Geometry

BEHS, Mrs. Prescott

Properties of Proportions from Section 8.1

1.

2.

. then , Ifc

d

a

b

d

c

b

a

Additional Properties of Proportions:

3. If

4. If

Ex. ¾ =9/12, then 3/9=4/12

Ex. ¾ = 9/12, then

(3+4)/4 = (9+12)/127/4 =

21/12

Examples:

True or False.

1.

2.

Complete the statement.

3.

4. .

15

23

9 then ,

15

8

9 If m

True

False

15

m+9

Example – page 469

26. 2

7

5

9 xX + 5

-7x -7x

K J

P

S

Q

Geometry Mean - definition

The geometric mean of 2 positive numbers a and b is the positive number x such that

Geometry Mean-Example

Find the geometric mean of 3 and 48.

√ √

Geometry Mean-Example

Find the geometric mean between 6 and 15.

Section 8.3 Similar Polygons

Page 473232 Geometry

BEHS, Mrs. Prescott

Madeleine Wood

Similar Polygons

• Definition: Similar Polygons – 2 polygons such that:

1. their corresponding angles are congruent2. the lengths of their corresponding sides are

proportional• Symbol:

Similar Polygons Example:1. You can see that the corresponding angles are

congruent.2. Corresponding sides are in proportion means that

the ratios of every 2 pairs of corresponding sides are equal.

m

m

m

m

m

m

m

m

4

2

6

3

2

1

10

5

They all reduce to ½ ,

which is called the

scale factor.

Similarity Statement: ABCD~EFGHThe order of the vertices indicates which angles correspond and which segments correspond in the similarity statement.To write a proportionality statement, write the ratios of all pairs of corresponding sides.

List all pairs of congruent corresponding angles.

Write the proportionality statement.

Examples: Page 475

2.

3.

15

83

10

Rotate to make it easier to match the corresponding sides

No, because 15/10 ≠ 8/3

Yes, all corr. ∡s are ≅ and the ratios of all corr. sides are =.

Examples: Page 475. Given: TUVW~ABCD

4.

5.

∡A≅∡T, ∡B≅∡U, ∡C≅∡V, ∡D≅∡W, andTW

AD

VW

CD

UV

BC

TU

AB

5

3

15

9

70º

A

D

B

T

C

U

VW

D

9

6

23

15

Scale factor of ABCD to TUVW

Scale factor of TUVW to ABCD is 5/3

Examples: Page 475

6.

7. Find all missing segment lengths and angle measures.

5

3

TW

AD

5

36x

10

303

x

x

110TUVm70º

A

D

B

T

C

U

V70º

)70180(TUVm

W

9

6

15

23

6

70º

70º

x =10 10

y

5

3

23

y

8.13

695

y

y

110º

110º 110º

110º

Given that ΔRST~ΔJKL. find the value of x and y.

x

5

7

4

354 x75.8x

2

6

7

4

y

42)2(4 y

5.8y

4284 y

50º

70º

(2w-5)º

70º

70 + 50 + (2w – 5) = 180115 + 2w = 180

2w = 65w = 32.5

344 y

Complete each.1. Find the scale factor of the

triangles.

2. Find the lengths of the missing segment lengths.

3. Find the perimeter of each triangle.

4. Find the ratio of the perimeter of the 2 triangles.

4

1

12

3

20,16 XZYZ

48201612

12543

4

1

20

12

Similar Polygon Perimeter TheoremIf 2 polygons are similar, then the ratio of their

perimeters is equal to the ratios of their corresponding side lengths.

MNONPOMP

ABCBDCAD

MN

AB

ON

CB

PO

DC

MP

AD

thenMPONADCBIf

,~

SECTION 8.4

1. Write the statement of proportionality.

2. m∡RSV=_______, m∡U= _______3. RS = ______ RT=______ SV= ______ RV=______

Given that ΔRSV~ΔRTU, answer each of the following.

R

S

T

V

U70º

80º5

10

8

12

𝑅𝑆𝑅𝑇

=𝑅𝑉𝑅𝑈

=𝑆𝑉𝑇𝑈

70˚ 80˚

5 15

4 4

• If 2 angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.

•Given that ∡D ≅ ∡A and ∡C ≅ ∡F, then ΔABC ~ ΔDEF.

AA ~ Postulate

A

BCF

D

E

Chapter 8- Section 8.5Proving Triangles are Similar

Page 488232 GeometryMrs. Prescott

BEHS

• If the lengths of the corresponding sides of two triangles are proportional, then the two triangles are similar.

SSS ~ Theorem

A

BCF

D

E

9

12

66

8

48

12

6

9

4

6

.~ then ,Given ABCDEFBC

EF

AC

DF

AB

DE

If an angle of one triangle is congruent to an angle of a 2nd triangle, and the lengths of the sides including these angle are proportional, then the two triangles are similar.

SAS ~ Theorem

A

BC

F

D

E

9

12

6

8

8

12

6

9,38,38 andBmEm

.~ then , andGiven ABCDEFBC

EF

AB

DEBE

38º

38º

1. AA~ Post., ΔABC ~ΔDEF

2. SAS~ Thm., ΔABC ~ΔDFE

3. Both ΔJKL and ΔMNP; SSS~

4. Ratio = 1:6, the triangles are similar by SSS~

Practice Problems: page 492 #2-5

Chapter 8- Section 8.6Proportions and Similar Triangles

Page 498232 GeometryMrs. Prescott

BEHS

A line that intersects 2 sides of a triangle is parallel to the 3rd side if and only if it divides the 2 sides proportionally.

Triangle Proportionality Theorem

R

S

T

V

U

5

10

8

4

.V then , If TUSVU

RV

ST

RS

. then ,V IfVU

RV

ST

RSTUS

and

Examples:

96

2 x

186 x3x

3

3

5

2x

103 x

3

10x

If 3 parallel lines intersect two transversals, then they divide the transversals proportionally.

Theorem:

AB

C

F

D

E

.,CE

BC

DF

ADthenFEDCABIf

Solve for x and y.Example:

84

3 x

84

5 y

6x

10y

• If a ray bisects an angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the other 2 sides.

Theorem:

F

D

E

9

12

3

4G

. DFEbisectsfFE

FD

GE

DGthenFGI

12

9

4

3

Find the value of A.Example:

A

6

8

5

485 A

6.9A