correctionkey=nl-c;ca-c name class date 12.4 similarity in...

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Explore Identifying Similarity in Right Triangles

A Make two copies of the right triangle on a piece of paper and cut them out.

B Choose one of the triangles. Fold the paper to find the altitude to the hypotenuse.

C Cut the second triangle along the altitude. Label the triangles as shown.

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Module 12 663 Lesson 4

12.4 Similarity in Right TrianglesEssential Question: How does the altitude to the hypotenuse of a right triangle help you

use similar right triangles to solve problems?

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D Place triangle 2 on top of triangle 1. What do you notice about the angles?

E What is true of triangles 1 and 2? How do you know?

F Repeat Steps 1 and 2 for triangles 1 and 3. Does the same relationship hold true for triangles 1 and 3?

Reflect

1. How are the hypotenuses of the triangles 2 and 3 related to triangle 1?

2. What is the relationship between triangles 2 and 3? Explain.

3. When you draw the altitude to the hypotenuse of a right triangle, what kinds of figures are produced?

4. Suppose you draw △ABC such that ∠B is a right angle and the altitude to the hypotenuse intersects hypotenuse

_ AC at point P. Match each triangle to a similar

triangle. Explain your reasoning.

A. △ABC △PAB

B. △PBC △CAB

C. △BAP △BPC



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Explain 1 Finding Geometric Means of Pairs of NumbersConsider the proportion a __ x = x __ b where two of the numbers in the proportion are the same. The number x is the geometric mean of a and b. The geometric mean of two positive numbers is the positive square root of their product. So the geometric mean of a and b is the positive number x such that x = √ ― ab or x 2 = ab.

Example 1 Find the geometric mean x of the numbers.

A 4 and 25

Write proportion. 4 _ x = x _ 25

Multiply both sides by the product of the denominators. 25x · 4 _ x = 25x · x _ 25

Multiply. 100x _ x = 25 x 2 _ 25

Simplify. 100 = x 2

Take the square root of both sides. √_

100 = √_

x 2

Simplify. 10 = x

B 9 and 20

Write proportion. _ x = x _ 20

Multiply both sides by the product of the denominators. 20x · _ x = 20x · x _ 20

Multiply. x _ x = 20 x 2 _ 20

Simplify. = x 2

Take the square root of both sides. √_

= √_

x 2

Simplify. = x

Reflect

5. How can you show that if positive numbers a and b are such that a __ x = x __ b , then x = √ ― ab ?

Your Turn

Find the geometric mean of the numbers. If necessary, give the answer in simplest radical form.

6. 6 and 24 7. 5 and 12


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Explain 2 Proving the Geometric Means TheoremsIn the Explore activity, you discovered a theorem about right triangles and similarity.

The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle.

That theorem leads to two additional theorems about right triangles. Both of the theorems involve geometric means.

Geometric Means Theorems

Theorem Example Diagram

The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse.

h 2 = xy or h = √ ― xy

The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse adjacent to that leg.

a 2 = xc or a = √ ― xc b 2 = yc or b = √ ― yc

Example 2 Prove the first Geometric Means Theorem.

Given: Right triangle ABC with altitude _ BD

Prove: CD _ BD = BD _ AD

Statements Reasons

1. △ABC with altitude _ BD 1. Given

2. △CBD ∼ △BAD 2.

3. CD _ BD

= BD _ AD

3.

Reflect

8. Discussion How can you prove the second Geometric Means Theorem?



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Explain 3 Using the Geometric Means TheoremsYou can use the Geometric Means Theorems to find unknown segment lengths in a right triangle.

Example 3 Find the indicated value.

A x

Write proportion. 2 _ x = x _ 10

Multiply both sides by the product of the denominators. 10x · 2 _ x = 10x · x _ 10

Multiply. 20x _ x = 10 x 2 _ 10

Simplify. 20 = x 2

Take the square root of both sides. √ ― 20 = √ ― x 2

Simplify. 2 √ ― 5 = x

B y

Write proportion. 10 _ y = y _ 12

Multiply both sides by the product of the denominators. 10 _ y = y _ 12

Multiply. _ y = _ 12

Simplify. =

Take the square root of both sides. √ ――― = √

――

Simplify. = y

Reflect

9. Discussion How can you check your answers?

Your Turn

10. Find x.



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Explain 4 Proving the Pythagorean Theorem using Similarity

You have used the Pythagorean Theorem in earlier courses as well as in this one. There are many, many proofs of the Pythagorean Theorem. You will prove it now using similar right triangles.

The Pythagorean Theorem

In a right triangle, the square of the sum of the lengths of the legs is equal to the square of the length of the hypotenuse.

Example 4 Complete the proof of the Pythagorean Theorem.

Given: Right △ABC

Prove: a 2 + b 2 = c 2

Part 1

Draw the altitude to the hypotenuse. Label the point of intersection X.

∠BXC ≅ ∠BCA because .

∠B ≅ ∠B by .

So, △BXC ∼ △BCA by .

∠AXC ≅ ∠ACB because .

∠A ≅ ∠A by .

So, △AXC ∼ △ACB by .

Part 2

Let the lengths of the segments of the hypotenuse be d and e, as shown in the figure.

Use the fact that corresponding sides of similar triangles are proportional to write two proportions.

Proportion 1: △BXC ∼ △BCA, so a _ c = _ a .

Proportion 2: △AXC ∼ △ACB, so b _ c = _____ b .



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Part 3

Now perform some algebra to complete the proof as follows.

Multiply both sides of Proportion1 by ac. Write the resulting equation.

Multiply both sides of Proportion12 by bc. Write the resulting equation.

Adding the two resulting equations give this:

Factor the right side of the equation:

Finally, use the fact that e + d = by the Segment Addition Postulate to rewrite the equation

as .

Reflect

11. Error Analysis A student used the figure in Part 2 of the example, and wrote the following incorrect proof of the Pythagorean Theorem. Critique the student’s proof. △BXC ∼ △BCA and △BCA ∼ △CXA, so △BXC ∼ △CXA by transitivity of similarity. Let CX = f. Since corresponding

sides of similar triangles are proportional, e _ f =

f _

d and f 2 = ed. Because △BXC ∼ △CXA and they

are right triangles, a 2 = e 2 + f 2 and b 2 = f 2 + d 2 .

Add the equations. a 2 + b 2 = e 2 + 2 f 2 + d 2

Substitute. = e 2 + 2ed + d 2

Factor. = (e + d) 2

Segment Addition Postulate = c 2



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Elaborate

12. How would you explain to a friend how to find the geometric mean of two numbers?

13. △XYZ is an isosceles right triangle and the right angle is ∠Y. Suppose the altitude to hypotenuse _ XZ

intersects _ XZ at point P. Describe the relationships among triangles △XYZ, △YPZ and △XPY.

14. Can two different pairs of numbers have the same geometric mean? If so, give an example. If not, explain why not.

15. Essential Question Check-In How is the altitude to the hypotenuse of a right triangle related to the segments of the hypotenuse it creates?



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• Online Homework• Hints and Help• Extra Practice

Evaluate: Homework and Practice

Write a similarity statement comparing the three triangles to each diagram.

1. 2. 3.

Find the geometric mean x of each pair of numbers. If necessary, give the answer in simplest radical form.

4. 5 and 20 5. 3 and 12

6. 8 and 13 7. 3.5 and 20

8. 1.5 and 84 9. 2 _ 3 and 27 _ 40

Find x, y, and z.

10. 11.



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12.

Use the diagram to complete each equation.

13. c _ e = _ d

14. c _ a = a _

15. c + d _ b

= b _ 16. d _ = e _ c

17. c (c + d) = 2

18. 2

= cd

Find the length of the altitude to the hypotenuse under the given conditions.

19. BC = 5 AC = 4

20. BC = 17 AC = 15

21. BC = 13 AC = 12



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22. Communicate Mathematical Ideas The area of a rectangle with a length of ℓ and a width of w has the same area as a square. Show that the side length of the square is the geometric mean of the length and width of the rectangle.

H.O.T. Focus on Higher Order Thinking

23. Algebra An 8-inch-long altitude of a right triangle divides the hypotenuse into two segments. One segment is 4 times as long as the other. What are the lengths of the segments of the hypotenuse?

24. Error Analysis Cecile and Amelia both found a value for EF in △DEF. Both students work are shown. Which student’s solution is correct? What mistake did the other student make?

Cecile: 12 ___ EF = EF ___ 8

So E F 2 = 12 (8) = 96.

Then EF = √_

96 = 4 √_

6 .

Amelia: 8 ___ EF = EF ___ 4

So E F 2 = 8 (4) = 32.

Then EF = √_

32 = 4 √_

2 .



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Lesson Performance TaskIn the example at the beginning of the lesson, a $100 investment grew for one year at the rate of 50%, to $150, then fell for one year at the rate of 50%, to $75. The arithmetic mean of +50% and -50%, which is 0%, was not a good predictor of the change, for it predicted the investment would still be worth $100 after two years, not $75.

1. Find the geometric mean of 1 + 50% and 1 - 50%. (Each 1 represents the fact that at the beginning of each year, an investment is worth 100% of itself.) Round to the nearest thousandth.

2. It is the geometric mean, not the arithmetic mean, that tells you what the interest rate would have had to have been over an entire investment period to achieve the end result. You can use your answer to Exercise 1 to check this claim. Find the value of a $100 investment after it increased or decreased at the rate you found in Exercise 1 for two years. Show your work.

3. Copy the right triangle shown here. Write the terms “Year 1 Rate”, “Year 2 Rate”, and “Average Rate” to show geometrically how the three investment rates relate to each other.

4. The geometric mean of n numbers is the nth root of the product of the numbers. Find what the interest rate would have had to have been over 4 years to achieve the result of a $100 investment that grew 20% in Year 1 and 30% in Year 2, then lost 20% in Year 3 and 30% in Year 4. Show your work. Round your answer to the nearest tenth of a percent.



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