Holt Geometry 7-4 Applying Properties of Similar Triangles Warm Up Solve each proportion. 1. 2. 3....

Holt Geometry

7-4 Applying Properties of Similar Triangles

Warm UpSolve each proportion.

AB = 16 QR = 10.5

x = 21

Holt Geometry

7-4 Applying Properties of Similar Triangles7-4 Applying Properties of Similar Triangles

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Geometry

Use properties of similar triangles to find segment lengths.Apply proportionality and triangle angle bisector theorems.

Objectives

Holt Geometry

Artists use mathematical techniques to make two-dimensional paintings appear three-dimensional. The invention of perspective was based on the observation that far away objects look smaller andcloser objects look larger.

Mathematical theorems like the Triangle Proportionality Theorem are important in making perspective drawings.

Holt Geometry

Example 1: Finding the Length of a Segment

Find US.

Substitute 14 for RU, 4 for VT, and 10 for RV.

Cross Products Prop.US(10) = 56

Divide both sides by 10.

It is given that , so by

the Triangle Proportionality Theorem.

Holt Geometry

Check It Out! Example 1

Find PN.

Substitute in the given values.

Cross Products Prop.2PN = 15

PN = 7.5 Divide both sides by 2.

Use the Triangle Proportionality Theorem.

Holt Geometry

Example 2: Verifying Segments are Parallel

Verify that .

Since , by the Converse of the

Triangle Proportionality Theorem.

Holt Geometry

AC = 36 cm, and BC = 27 cm.

Verify that .

Since , by the Converse of the

Triangle Proportionality Theorem.

Holt Geometry

The previous theorems and corollary lead to the following conclusion.

Holt Geometry

Example 4: Using the Triangle Angle Bisector Theorem

Find PS and SR.

Substitute the given values.

Cross Products Property

Distributive Property

by the ∆ Bisector Theorem.

40(x – 2) = 32(x + 5)

40x – 80 = 32x + 160

Holt Geometry

Example 4 Continued

Simplify.

Substitute 30 for x.

40x – 80 = 32x + 160

8x = 240

x = 30

PS = x – 2 SR = x + 5

= 30 – 2 = 28 = 30 + 5 = 35

Holt Geometry

Find AC and DC.

Substitute in given values.

Cross Products Theorem

So DC = 9 and AC = 16.

Simplify.

by the ∆ Bisector Theorem.

4y = 4.5y – 9

–0.5y = –9

Divide both sides by –0.5. y = 18

Holt Geometry

Find the length of each segment.

Lesson Quiz: Part I

SR = 25, ST = 15

Holt Geometry

Lesson Quiz: Part II

3. Verify that BE and CD are parallel.

Since , by the

Converse of the ∆ Proportionality Thm.

7-5 Using Proportional Relationships7-5 Using Proportional Relationships

Holt Geometry

Warm UpWarm Up

7-5 Using Proportional Relationships

indirect measurementscale drawingscale

Vocabulary

Indirect measurement is any method that uses formulas, similar figures, and/or proportions to measure an object. The following example shows one indirect measurement technique.

Whenever dimensions are given in both feet and inches, you must convert them to either feet or inches before doing any calculations.

Helpful Hint

Example 1: Measurement Application

Tyler wants to find the height of a telephone pole. He measured the pole’s shadow and his own shadow and then made a diagram. What is the height h of the pole?

Example 1 Continued

Step 1 Convert the measurements to inches.

Step 2 Find similar triangles.

Because the sun’s rays are parallel, A F. Therefore ∆ABC ~ ∆FGH by AA ~.

AB = 7 ft 8 in. = (7 12) in. + 8 in. = 92 in.

BC = 5 ft 9 in. = (5 12) in. + 9 in. = 69 in.

FG = 38 ft 4 in. = (38 12) in. + 4 in. = 460 in.

Example 1 Continued

Step 3 Find h.

The height h of the pole is 345 inches, or 28 feet 9 inches.

Corr. sides are proportional.

Substitute 69 for BC, h for GH, 92 for AB, and 460 for FG.

Cross Products Prop.92h = 69 460

Divide both sides by 92.h = 345

A student who is 5 ft 6 in. tall measured shadows to find the height LM of a flagpole. What is LM?

Step 1 Convert the measurements to inches.

GH = 5 ft 6 in. = (5 12) in. + 6 in. = 66 in.

JH = 5 ft = (5 12) in. = 60 in.

NM = 14 ft 2 in. = (14 12) in. + 2 in. = 170 in.

Check It Out! Example 1 Continued

Step 2 Find similar triangles.

Because the sun’s rays are parallel, L G. Therefore ∆JGH ~ ∆NLM by AA ~.

Step 3 Find h.

Corr. sides are proportional.

Substitute 66 for BC, h for LM, 60 for JH, and 170 for MN.

Cross Products Prop.

60(h) = 66 170

h = 187

The height of the flagpole is 187 in., or 15 ft. 7 in.

A scale drawing represents an object as smaller than or larger than its actual size. The drawing’s scale is the ratio of any length in the drawingto the corresponding actual length. For example, on a map with a scale of 1 cm : 1500 m, one centimeter on the map represents 1500 m in actual distance.

A proportion may compare measurements that have different units.

Remember!

The rectangular central chamber of the Lincoln Memorial is 74 ft long and 60 ft wide. Make a scale drawing of the floor of the chamber using a scale of 1 in.:20 ft.

Set up proportions to find the length l and width w of the scale drawing.

20w = 60

w = 3 in

3.7 in.

Example 4: Using Ratios to Find Perimeters and Areas

Given that ∆LMN:∆QRT, find the perimeter P and area A of ∆QRS.

The similarity ratio of ∆LMN to

∆QRS is

By the Proportional Perimeters and Areas Theorem,

the ratio of the triangles’ perimeters is also , and

the ratio of the triangles’ areas is

Example 4 Continued

Perimeter Area

The perimeter of ∆QRS is 25.2 cm, and the area is 29.4 cm2.

13P = 36(9.1)

P = 25.2

132A = (9.1)2(60)

A = 29.4 cm2

∆ABC ~ ∆DEF, BC = 4 mm, and EF = 12 mm. If P = 42 mm and A = 96 mm2 for ∆DEF, find the perimeter and area of ∆ABC.

Perimeter Area

The perimeter of ∆ABC is 14 mm, and the area is 10.7 mm2.

12P = 42(4)

P = 14 mm

122A = (4)2(96)

Lesson Quiz: Part I

1. Maria is 4 ft 2 in. tall. To find the height of a flagpole, she measured her shadow and the pole’s shadow. What is the height h of the flagpole?

2. A blueprint for Latisha’s bedroom uses a scale of 1 in.:4 ft. Her bedroom on the blueprint is 3 in. long. How long is the actual room?

3. ∆ABC ~ ∆DEF. Find the perimeter and area of ∆ABC.

P = 27 in., A = 31.5 in2

Holt Geometry

7-6 Dilations and Similarity in the Coordinate Plane7-6 Dilations and Similarity

in the Coordinate Plane

Holt Geometry

Warm UpWarm Up

Holt Geometry

7-6 Dilations and Similarity in the Coordinate Plane

Warm UpSimplify each radical.

1. 2. 3.

Find the distance between each pair of points. Write your answer in simplest radical form.

4. C (1, 6) and D (–2, 0)

5. E(–7, –1) and F(–1, –5)

Holt Geometry

Apply similarity properties in the coordinate plane.

Use coordinate proof to prove figures similar.

Objectives

Holt Geometry

dilationscale factor

Vocabulary

Holt Geometry

If the scale factor of a dilation is greater than 1 (k > 1), it is an enlargement. If the scale factor is less than 1 (k < 1), it is a reduction.

Helpful Hint

Holt Geometry

7-6 Dilations and Similarity in the Coordinate PlaneExample 1: Computer Graphics Application

Draw the border of the photo after a

dilation with scale factor

Holt Geometry

Example 1 Continued

RectangleABCD

Step 1 Multiply the vertices of the photo A(0, 0), B(0,

4), C(3, 4), and D(3, 0) by

RectangleA’B’C’D’

Holt Geometry

Example 1 Continued

Step 2 Plot points A’(0, 0), B’(0, 10), C’(7.5, 10), and D’(7.5, 0).

Draw the rectangle.

Holt Geometry

What if…? Draw the border of the original photo

after a dilation with scale factor

Holt Geometry

Check It Out! Example 1 Continued Step 1 Multiply the vertices of the photo A(0, 0), B(0,

4), C(3, 4), and D(3, 0) by

RectangleABCD

RectangleA’B’C’D’

Holt Geometry

Step 2 Plot points A’(0, 0), B’(0, 2), C’(1.5, 2), and D’(1.5, 0).

Draw the rectangle.

A’ D’

B’ C’

Holt Geometry

Example 2: Finding Coordinates of Similar Triangle

Given that ∆TUO ~ ∆RSO, find the coordinates of U and the scale factor.

Since ∆TUO ~ ∆RSO,

Substitute 12 for RO, 9 for TO, and 16 for OY.

12OU = 144 Cross Products Prop.

OU = 12 Divide both sides by 12.

Holt Geometry

Example 2 Continued

U lies on the y-axis, so its x-coordinate is 0. Since OU = 12, its y-coordinate must be 12. The coordinates of U are (0, 12).

So the scale factor is

Holt Geometry

Example 3: Proving Triangles Are Similar

Prove: ∆EHJ ~ ∆EFG.

Given: E(–2, –6), F(–3, –2), G(2, –2), H(–4, 2), and J(6, 2).

Step 1 Plot the points and draw the triangles.

Holt Geometry

Example 3 Continued

Step 2 Use the Distance Formula to find the side lengths.

Holt Geometry

Example 3 Continued

Step 3 Find the similarity ratio.

= 2 = 2

Since and E E, by the Reflexive Property,

∆EHJ ~ ∆EFG by SAS ~ .

Holt Geometry

7-6 Dilations and Similarity in the Coordinate PlaneExample 4: Using the SSS Similarity Theorem

Verify that ∆A'B'C' ~ ∆ABC.

Graph the image of ∆ABC

after a dilation with scale

factor

Holt Geometry

Example 4 Continued

Step 1 Multiply each coordinate by to find the

coordinates of the vertices of ∆A’B’C’.

Holt Geometry

Step 2 Graph ∆A’B’C’.

Example 4 Continued

C’ (4, 0)

A’ (0, 2)

B’ (2, 4)

Holt Geometry

Example 4 Continued

Holt Geometry

Example 4 Continued

Since , ∆ABC ~ ∆A’B’C’ by SSS ~.

Holt Geometry

Graph the image of ∆MNP after a dilation with scale factor 3. Verify that ∆M'N'P' ~ ∆MNP.

Holt Geometry

Step 1 Multiply each coordinate by 3 to find the coordinates of the vertices of ∆M’N’P’.

Holt Geometry

Step 2 Graph ∆M’N’P’.

-7 -6 -5 -4 -3 -2 -1 1 2 3 4 5 6 7

-7-6-5-4-3-2-1

1234567

Holt Geometry

Since , ∆MNP ~ ∆M’N’P’ by SSS ~.

Holt Geometry

Lesson Quiz: Part I

1. Given X(0, 2), Y(–2, 2), and Z(–2, 0), find the

coordinates of X', Y, and Z' after a dilation with

scale factor –4.

2. ∆JOK ~ ∆LOM. Find the coordinates of M and the scale factor.

X'(0, –8); Y'(8, –8); Z'(8, 0)

Holt Geometry

3. Given: A(–1, 0), B(–4, 5), C(2, 2), D(2, –1),

E(–4, 9), and F(8, 3)

Prove: ∆ABC ~ ∆DEF

Therefore, and ∆ABC ~ ∆DEF

by SSS ~.

Holt Geometry 7-4 Applying Properties of Similar Triangles Warm Up Solve each proportion. 1. 2. 3....

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