1. Activator Solve the proportion: 4/ (x+2) = 16/(x + 5) Simplify:
5weeks/30days; 85cm/.5m

2. UNIT E.Q
How can I use all what I am learning on polygon similarity to solve real life problems.
Which careers use similarity most?
Is similarity shown and used only in geometry?

3. Standards for Mathematical Practice
1. Make sense of problems and persevere in solving them.
2. Reason abstractly and quantitatively.
3. Construct viable arguments and critique the reasoning of others.  
4. Model with mathematics.
5. Use appropriate tools strategically.
6. Attend to precision.
7. Look for and make use of structure.
8. Look for and express regularity in repeated reasoning.

4. 8.1 Ratio and Proportion

5. Essential Question: 1. How do we use proportions to solve problems?
2.How do we use properties of proportions to solve real- life problems, such as using the scale of a map?

6. Computing Ratios Ratio of a to b : Ex: Simplify the ratio of 6 to 8.

7. Simplifying Ratios
Simplify the ratios.

8. The perimeter of rectangle ABCD is 60 cm. The ratio of AB:BC is 3:2
The perimeter of rectangle ABCD is 60 cm. The ratio of AB:BC is 3:2. Find the length and width of the rectangle.
Hint: draw a rectangle ABCD.

9. Using Extended Ratios
The measure of the angles in ∆JKL are in the extended ratio of 1:2:3. Find the measures of the angles. 2x 3x x

10. Using Ratios
The ratios of the side lengths of ∆DEF to the corresponding side lengths of ∆ABC are 2:1. Find the unknown lengths. A D 3 L B C L F E 8

11. Properties of proportions
Cross product property
then ad = bc.
Reciprocal property
Introduce means and extremes

12. Solve the proportions.

13. Given that the track team won 8 meets and lost 2, find the ratios.
What is the ratio of wins to loses?
What is the ratio of losses to wins?
What is the ratio of wins to the total number of track meets?

14. Simplify the ratio. 3 ft to 12 in 60 cm to 1 m 350g to 1 kg
6 meters to 9 meters

15. Solve the proportion

16. Tell whether the statement is true.

17. 8.2 Problem Solving in Geometry with Proportions

18. Additional Properties of Proportions
If , then
If , then

19. Using Properties of Proportions
Tell whether the statement is true.

20. In the diagram
Find the length of BD. A 30 16 B C x 10 D E

21. In the diagram
Solve for DE. A 5 2 D B 3 E C

22. Geometric mean
The geometric mean of two positive numbers a and b is the positive number x such that
Find the geometric mean of 8 and 18.

23. Geometric Mean Find the geometric mean of 5 and 20.
The geometric mean of x and 5 is 15. Find the value of x.

24. Different perspective of Geometric mean
The geometric mean of ‘a’ and ‘b’ is √ab
Therefore geometric mean of 4 and 9 is 6, since √(4)(9) = √36 = 6.

25. Geometric mean Find the geometric mean of the two numbers. 3 and 27
√(3)(27) = √81 = 9
4 and 16
√(4)(16) = √64 = 8
5 and 15
√(5)(15) = √75 = 5√3

26. Ex. 3: Using a geometric mean
PAPER SIZES. International standard paper sizes are commonly used all over the world. The various sizes all have the same width-to-length ratios. Two sizes of paper are shown, called A4 and A3. The distance labeled x is the geometric mean of 210 mm and 420 mm. Find the value of x.
Slide #26

27. Write proportion 210 x = x 420 X2 = 210 ∙ 420 X = √210 ∙ 420
Solution:
The geometric mean of 210 and 420 is 210√2, or about 297mm.
210 x
Write proportion = x
420
X2 = 210 ∙ 420
X = √210 ∙ 420
X = √210 ∙ 210 ∙ 2
X = 210√2
Cross product property
Simplify
Factor
Simplify
Slide #27

28. Using proportions in real life
In general when solving word problems that involve proportions, there is more than one correct way to set up the proportion.
Slide #28

29. Ex. 4: Solving a proportion
MODEL BUILDING. A scale model of the Titanic is inches long and inches wide. The Titanic itself was feet long. How wide was it?
Width of Titanic
Length of Titanic =
Width of model
Length of model
LABELS:
Width of Titanic = x
Width of model ship = in
Length of Titanic = feet
Length of model ship = in.
Slide #29

30. Reasoning:
Write the proportion. Substitute. Multiply each side by Use a calculator.
Width of Titanic
Length of Titanic =
Width of model
Length of model
x feet
feet =
11.25 in.
107.5 in.
11.25(882.75) x =
107.5 in.
x ≈ 92.4 feet
So, the Titanic was about 92.4 feet wide.
Slide #30

31. Note:
Notice that the proportion in this Example contains measurements that are not in the same units. When writing a proportion in unlike units, the numerators should have the same units and the denominators should have the same units.
The inches (units) cross out when you cross multiply.
Slide #31

32. A model truck is 13. 5 inches long and 7. 5 inches wide
A model truck is 13.5 inches long and 7.5 inches wide. The original truck was 12 feet long. How wide was it?

33. 8.3 Similar Polygons

34. Activator

35. Identifying Similar Polygons
When there is a correspondence between two polygons such that their corresponding angles are congruent and the lengths of corresponding sides are proportional the two polygons are called similar polygons.

36. Similar polygons
If ABCD ~ EFGH,
then G H C D B F A E

37. Similar polygons Given ABCD ~ EFGH, solve for x. 2x = 24 x = 12 6 x 2

38. Is ABC ~ DEF? Explain. B D 6 E 13 12 5 7 A C F 10
ABC is not similar to DEF since corresponding sides are not proportional. ? ?
yes no

39. Similar polygons Given ABCD ~ EFGH, solve for the variables. 5 x 10 26 A E

40. Ex: Scale factor of this triangle is 1:2
If two polygons are similar, then the ratio of the lengths of two corresponding sides is called the scale factor.
Ex: Scale factor of this triangle is 1:2
4.5 9 3 6

41. Quadrilateral JKLM is similar to PQRS. Find the value of z.15 Q z 6 P J M 10
15z = 60
z = 4

42. Theorem
If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths.
If KLMN ~ PQRS, then

43. Given ABC ~ DEF, find the scale factor of ABC to DEF and find the perimeter of each polygon.
= 40 B
P =
= 20 12 20 6 10 A C D F 4 8
CORRESPONDING SIDES
4 : 8
1 : 2

45. 8.4 Similar Triangles

46. In the diagram, ∆BTW ~ ∆ETC.
Write the statement of proportionality.
Find m<TEC.
Find ET and BE. T
34° E C 3 20
79° B W 12

47. Postulate 25 Angle-Angle (AA) Similarity Postulate
If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.

48. Similar Triangles
Given the triangles are similar. Find the value of the variable. )) m )) ) 6 8
11m = 48 ) 11

49. Similar Triangles
Given the triangles are similar. Find the value of the variable.
Left side of sm Δ Base of sm Δ
Left side of lg Δ Base of lg Δ = 6 5
> 2
6h = 40
> h

50. ∆ABC ≈ ∆DBE. A 5 D y 9 x B C 8 E 4

51. Determine whether the triangles are similar.6
32°
33° 9 18
No, because two angles of one triangle are not congruent
to two angles of another triangle.

52. Determine whether the triangles are similar.
60°
60°
60°
60°
Yes, because two angles of one triangle are congruent
to two angles of another triangle.

53. Given two triangles are similar, solve for the variables.14 16 15 ) ) 10
15(a+3) = 10(16)
15a + 45 = 160
15a = 115

54. Decide whether two triangles are similar, not similar, or cannot be determined.
92°
31°
47° S
41°
92°
57°
S = 180
S = 180
S = 47
A = 180
A = 180
A = 31
Not similar

55. 8.5 Proving Triangles are Similar

56. Objectives:
Use similarity theorems to prove that two triangles are similar
Use similar triangles to solve real-life problems such as finding the height of a climbing wall.

57. Using Similarity Theorems
In this lesson, you will study 2 alternate ways of proving that two triangles are similar: Side-Side-Side Similarity Theorem and the Side-Angle-Side Similarity Theorem. The first theorem is proved in Example 1 and you are asked to prove the second in Exercise 31.

58. Side Side Side(SSS) Similarity Theorem
If the corresponding sides of two triangles are proportional, then the triangles are similar.
THEN ∆ABC ~ ∆PQR AB BC CA = = PQ QR RP

59. Side Angle Side Similarity Thm.
If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar. ZX XY
If X  M and = PM MN
THEN ∆XYZ ~ ∆MNP

60. Ex. 1: Proof of Theorem 8.2 Given: ProveRS ST TR
∆RST ~ ∆LMN = = LM MN NL
Locate P on RS so that PS = LM. Draw PQ so that PQ ║ RT. Then ∆RST ~ ∆PSQ, by the AA Similarity Postulate, and RS ST TR = = LM MN NL
Because PS = LM, you can substitute in the given proportion and find that SQ = MN and QP = NL. By the SSS Congruence Theorem, it follows that ∆PSQ  ∆LMN Finally, use the definition of congruent triangles and the AA Similarity Postulate to conclude that ∆RST ~ ∆LMN.

61. Ex. 2: Using the SSS Similarity Thm.
Which of the three triangles are similar?
To decide which, if any, of the triangles are similar, you need to consider the ratios of the lengths of corresponding sides.
Ratios of Side Lengths of ∆ABC and ∆DEF. AB 6 3 CA 12 3 BC 9 3 = = = = = = DE 4 2 FD 8 2 EF 6 2
Because all of the ratios are equal, ∆ABC ~ ∆DEF.

62. Ratios of Side Lengths of ∆ABC ~ ∆GHJ6 CA 12 6 BC 9 = = 1 = = = GH 6 JG 14 7 HJ 10
Because the ratios are not equal, ∆ABC and ∆GHJ are not similar.
Since ∆ABC is similar to ∆DEF and ∆ABC is not similar to ∆GHJ, ∆DEF is not similar to ∆GHJ.

63. Ex. 3: Using the SAS Similarity Thm.
Use the given lengths to prove that ∆RST ~ ∆PSQ.
Given: SP=4, PR = 12, SQ = 5, and QT = 15;
Prove: ∆RST ~ ∆PSQ
Use the SAS Similarity
Theorem. Begin by finding the ratios of the lengths of the corresponding sides. SR
SP + PR
4 + 12 16 = 4 = = = SP SP 4 4

64. ST
SQ + QT
5 + 15 20 = 4 = = = SQ SQ 5 5
So, the side lengths SR and ST are proportional to the corresponding side lengths of ∆PSQ. Because S is the included angle in both triangles, use the SAS Similarity Theorem to conclude that ∆RST ~ ∆PSQ.

65. Using Similar Triangles in Real Life
Ex. 6 – Finding Distance indirectly.
To measure the width of a river, you use a surveying technique, as shown in the diagram.

66. Solution By the AA Similarity Postulate, ∆PQR ~ ∆STR.RQ PQ =
Write the proportion. RT ST RQ 63 =
Substitute. 12 9 RQ =
12 ● 7
Multiply each side by 12. RQ = 84
Solve for TS.
So the river is 84 feet wide.

67. 8.6 Proportions & Similar Triangles

68. Objectives: Use proportionality theorems to calculate segment lengths.
To solve real-life problems, such as determining the dimensions of a piece of land.

69. Use Proportionality Theorems
In this lesson, you will study four proportionality theorems. Similar triangles are used to prove each theorem.

70. Theorems 8.4 Triangle Proportionality Theorem
If a line parallel to one side of a triangle intersects the other two sides, then it divides the two side proportionally.
If TU ║ QS, then RT RU = TQ US

71. Theorems 8.5 Converse of the Triangle Proportionality Theorem
If a line divides two sides of a triangle proportionally, then it is parallel to the third side. RT RU If
, then TU ║ QS. = TQ US

72. Ex. 1: Finding the length of a segment
In the diagram AB ║ ED, BD = 8, DC = 4, and AE = 12. What is the length of EC?

73. Triangle Proportionality Thm.
Step:
DC EC
BD AE
4 EC
4(12) 8
6 = EC
Reason
Triangle Proportionality Thm.
Substitute
Multiply each side by 12.
Simplify. = = EC =
So, the length of EC is 6.

74. Ex. 2: Determining Parallels
Given the diagram, determine whether MN ║ GH. LM 56 8 = = MG 21 3 LN 48 3 = = NH 16 1 8 3 ≠ 3 1
MN is not parallel to GH.

75. Theorem 8.6
If three parallel lines intersect two transversals, then they divide the transversals proportionally.
If r ║ s and s║ t and l and m intersect, r, s, and t, then UW VX = WY XZ

76. Theorem 8.7
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 two sides.
If CD bisects ACB, then AD CA = DB CB

77. Ex. 3: Using Proportionality Theorems
In the diagram 1  2  3, and PQ = 9, QR = 15, and ST = 11. What is the length of TU?

78. 9 ● TU = 15 ● 11 Cross Product property
SOLUTION: Because corresponding angles are congruent, the lines are parallel and you can use Theorem 8.6 PQ ST
Parallel lines divide transversals proportionally. = QR TU 9 11 =
Substitute 15 TU
9 ● TU = 15 ● 11 Cross Product property
15(11) 55 TU = =
Divide each side by 9 and simplify. 9 3
So, the length of TU is 55/3 or 18 1/3.

79. Ex. 4: Using the Proportionality Theorem
In the diagram, CAD  DAB. Use the given side lengths to find the length of DC.

80. Solution:
Since AD is an angle bisector of CAB, you can apply Theorem Let x = DC. Then BD = 14 – x. AB BD =
Apply Thm. 8.7 AC DC 9
14-X
Substitute. = 15 X

81. Ex. 4 Continued . . . 9 ● x = 15 (14 – x) Cross product property
Distributive Property
Add 15x to each side
Divide each side by 24.
So, the length of DC is 8.75 units.

82. Use proportionality Theorems in Real Life
Example 5: Finding the length of a segment
Building Construction: You are insulating your attic, as shown. The vertical 2 x 4 studs are evenly spaced. Explain why the diagonal cuts at the tops of the strips of insulation should have the same length.

83. Use proportionality Theorems in Real Life
Because the studs AD, BE and CF are each vertical, you know they are parallel to each other. Using Theorem 8.6, you can conclude that DE AB = EF BC
Because the studs are evenly spaced, you know that DE = EF. So you can conclude that AB = BC, which means that the diagonal cuts at the tops of the strips have the same lengths.

84. Ex. 6: Finding Segment Lengths
In the diagram KL ║ MN. Find the values of the variables.

85. Solution To find the value of x, you can set up a proportion.9 x =
Write the proportion
Cross product property
Distributive property
Add 13.5x to each side.
Divide each side by 22.5
13.5 x
13.5(37.5 – x) = 9x
– 13.5x = 9x
= 22.5 x
22.5 = x
Since KL ║MN, ∆JKL ~ ∆JMN and JK KL = JM MN

86. Solution To find the value of y, you can set up a proportion.9
7.5 =
Write the proportion
Cross product property
Divide each side by 9. y
9y = 7.5(22.5)
y = 18.75

87. 8.7 Dilations

88. Objectives Identify dilations
Use properties of dilations to create a real-life perspective drawing.

89. Identifying Dilations
In chapter 7, you studied rigid transformations, in which the image and preimage of a figure are congruent. In this lesson, you will study a type of nonrigid transformation called a dilation, in which the image and preimage are similar.

90. What is it?
A dilation with center C and a scale factor k is a transformation that maps every point P in the plane to a point P’ so that the following properties are true.
If P is not the center point C, then the image point P’ lies on CP. The scale factor k is a positive number such that k =
and k ≠1.
2. If P is the center point C, then P = P’.
CP’ CP

91. Reduction/Enlargement
The dilation is a reduction if 0 < k < 1 and it is an enlargement if k > 1.
CP’ 3 1
REDUCTION: = = CP 6 2 6

92. Is equal to the scale factor of the dilation.
CP’ 5 =
ENLARGEMENT: CP 2 5
Because ∆PQR ~ ∆P’Q’R’
P’Q’
Is equal to the scale factor of the dilation. PQ

93. Ex. 1: Identifying Dilations
Identify the dilation and find its scale factor.
CP’ 2 =
REDUCTION: CP 3 2
The scale factor is k = This is a reduction. 3

94. Ex. 1B -- Enlargement Identify the dilation and find its scale factor.
CP’ 2 = 2 =
ENLARGEMENT: CP 1 2
The scale factor is k = This is an enlargement. = 2 1

95. Notes:
In a coordinate plane, dilations whose centers are the origin have the property that the image of P (x, y) is P’ (kx, ky)

96. Ex. 2: Dilation in a coordinate plane
Draw a dilation of rectangle ABCD with A(2, 2), B(6, 2), C(6, 4), and D(2, 4). Use the origin as the center and use a scale factor of ½. How does the perimeter of the preimage compare to the perimeter of the image?

97. SOLUTION:
Because the center of the dilation is the origin, you can find the image of each vertex by multiplying is coordinates by the scale factor
A(2, 2) A’(1, 1)
B(6, 2) B’(3, 1)
C(6, 4) C’(3, 2)
D(2, 4) D’(1, 2)

98. Solution continued
From the graph, you can see that the preimage has a perimeter of 12 and the image has a perimeter of 6. A preimage and its image after a dilation are similar figures. Therefore, the ratio of perimeters of a preimage and its image is equal to the scale factor of the dilation.

99. Using Dilations in Real Life—p.508
Finding Scale Factor: Shadow puppets have been used in many countries for hundreds of years. A flat figure is held between a light and a screen. The audience on the other side of the screen sees the puppet’s shadow. The shadow is a dilation, or enlargement of the shadow puppet. When looking at a cross sectional view, ∆LCP ~ ∆LSH.

100. Shadow Puppet continued
The shadow puppet shown is 12 inches tall. (CP in the diagram). Find the height of the shadow, SH, for each distance from the screen. In each case, by what percent is the shadow larger than the puppet?
A. LC = LP = 59 in.; LS = LH = 74 in.
B. LC = LP = 66 in.; LS = LH = 74 in.

101. Finding Scale Factor So, the shadow is 25% larger than the puppet. 5912 LC CP =
ENLARGEMENT: = 74 SH LS SH
59SH = 75(12)
59SH = 888
SH ≈15 INCHES
To find the percent of the size increase, use the scale factor of the dilation. SH
Scale factor = CP 15
1.25 = 12
So, the shadow is 25% larger than the puppet.

102. Finding Scale Factor
Notice that as the puppet moves closer to the screen, the shadow height increase. 66 12 LC CP =
ENLARGEMENT: = 74 SH LS SH
66SH = 75(12)
66SH = 888
SH ≈13.45 INCHES
Use the scale factor again to find the percent of size increase. SH
Scale factor = CP
13.45
1.12 = 12
So, the shadow is 12% larger than the puppet.