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# 8.3: Using Similar Right Triangles

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## Learning Objectives

• Identify similar triangles inscribed in a larger triangle.
• Use proportions in similar right triangles.

## Review Queue

1. Solve the following ratios.
1. $\frac{3}{x} = \frac{x}{27}$
2. $\frac{ \sqrt{6} }{x} = \frac{x}{9 \sqrt{6}}$
3. $\frac{x}{15} = \frac{12}{x}$
2. If the legs of an isosceles right triangle are 4, find the length of the hypotenuse. Draw a picture and simplify the radical.

Know What? The bridge to the right is called a truss bridge. It is a steel bridge with a series of right triangles that are connected as support. All the red right triangles are similar. Can you find $x, y$ and $z$?

## Inscribed Similar Triangles

You may recall that if two objects are similar, corresponding angles are congruent and their sides are proportional in length.

Theorem 8-5: If an altitude is drawn from the right angle of any right triangle, then the two triangles formed are similar to the original triangle and all three triangles are similar to each other.

In $\triangle ADB, m \angle A = 90^{\circ}$ and $\overline{AC} \bot \overline{DB}$, then $\triangle ADB \sim \triangle CDA \sim \triangle CAB$.

Example 1: Write the similarity statement for the triangles below.

Solution: Separate out the three triangles.

Line up the congruent angles: $\triangle IRE \sim \triangle ITR \sim \triangle RTE$

We can also use the side proportions to find the length of the altitude.

Example 2: Find the value of $x$.

Solution: Separate the triangles to find the corresponding sides.

Set up a proportion.

$\frac{\text{shorter leg in} \ \triangle EDG}{\text{shorter leg in} \ \triangle DFG} &= \frac{\text{hypotenuse in} \ \triangle EDG}{\text{hypotenuse in} \ \triangle DFG}\\\frac{6}{x} &= \frac{10}{8}\\48 &= 10x\\4.8 &= x$

Example 3: Find the value of $x$.

Solution: Set up a proportion.

$\frac{\text{shorter leg in} \ \triangle SVT}{\text{shorter leg in} \ \triangle RST} &= \frac{\text{hypotenuse in} \ \triangle SVT}{\text{hypotenuse in} \ \triangle RST}\\\frac{4}{x} &= \frac{x}{20}\\x^2 &= 80\\x &= \sqrt{80} = 4 \sqrt{5}$

Example 4: Find the value of $y$ in $\triangle RST$ above.

Solution: Use the Pythagorean Theorem.

$y^2 + \left( 4 \sqrt{5} \right )^2 &= 20^2\\y^2 + 80 &= 400\\y^2 &= 320\\y &= \sqrt{320} = 8 \sqrt{5}$

## The Geometric Mean

Geometric Mean: The geometric mean of two positive numbers $a$ and $b$ is the positive number $x$, such that $\frac{a}{x} = \frac{x}{b}$ or $x^2 = ab$ and $x = \sqrt{ab}$.

Example 5: Find the geometric mean of 24 and 36.

Solution: $x = \sqrt{24 \cdot 36} = \sqrt{12 \cdot 2 \cdot 12 \cdot 3} = 12 \sqrt{6}$

Example 6: Find the geometric mean of 18 and 54.

Solution: $x = \sqrt{18 \cdot 54} = \sqrt{18 \cdot 18 \cdot 3} = 18 \sqrt{3}$

In both of these examples, we did not multiply the numbers together. This makes it easier to simplify the radical. A practical application of the geometric mean is to find the altitude of a right triangle.

Example 7: Find the value of $x$.

Solution: Set up a proportion.

$\frac{shortest \ leg \ of \ smallest \ \triangle}{shortest \ leg \ of \ middle \ \triangle} &= \frac{longer \ leg \ of \ smallest \ \triangle}{longer \ leg \ of \ middle \ \triangle}\\\frac{9}{x} &= \frac{x}{27}\\x^2 &= 243\\x &= \sqrt{243} = 9 \sqrt{3}$

In Example 7, $\frac{9}{x} = \frac{x}{27}$ is in the definition of the geometric mean. So, the altitude is the geometric mean of the two segments that it divides the hypotenuse into. In other words, $\frac{BC}{AC} = \frac{AC}{DC}$. Two other true proportions are $\frac{BC}{AB} = \frac{AB}{DB}$ and $\frac{DC}{AD} = \frac{AD}{DB}$.

Example 8: Find the value of $x$ and $y$.

Solution: Separate the triangles. Write a proportion for $x$.

$\frac{20}{x} &= \frac{x}{35}\\ x^2 &= 20 \cdot 35\\x &= \sqrt{20 \cdot 35}\\x &= 10 \sqrt{7}$

Set up a proportion for $y$. Or, you can use the Pythagorean Theorem to solve for $y$.

$\frac{15}{y} &= \frac{y}{35} && (10 \sqrt{7})^2 + y^2 = 35^2\\y^2 &= 15 \cdot 35 && \qquad 700 + y^2 =1225\\y &= \sqrt{15 \cdot 35} && \qquad \qquad \quad \ y = \sqrt{525} = 5 \sqrt{21}\\y &= 5 \sqrt{21} && \text{Use the method you feel most comfortable with.}$

Know What? Revisited To find the hypotenuse of the smallest triangle, do the Pythagorean Theorem.

$45^2 + 28^2 &= x^2\\2809 &= x^2\\53 &= x$

Because the triangles are similar, find the scale factor of $\frac{70}{28} = 2.5$.

$y = 45 \cdot 2.5 = 112.5$ and $z=53 \cdot 2.5=135.5$

## Review Questions

• Questions 1-4 use the ratios of similar right triangles.
• Questions 5-8 are similar to Example 1.
• Questions 9-11 are similar to Examples 2-4
• Questions 12-17 are similar to Examples 5 and 6.
• Questions 18-29 are similar to Examples 2, 3, 4, 7, and 8.
• Question 30 is a proof of theorem 8-5.

Fill in the blanks.

1. $\triangle BAD \sim \triangle \underline{\;\;\;\;\;\;\;\;\;} \sim \triangle \underline{\;\;\;\;\;\;\;\;\;}$
2. $\frac{BC}{?} = \frac{?}{CD}$
3. $\frac{BC}{AB} = \frac{AB}{?}$
4. $\frac{?}{AD} = \frac{AD}{BD}$

Write the similarity statement for the right triangles in each diagram.

Use the diagram to answer questions 8-11.

1. Write the similarity statement for the three triangles in the diagram.
2. If $JM = 12$ and $ML = 9$, find $KM$.
3. Find $JK$.
4. Find $KL$.

Find the geometric mean between the following two numbers. Simplify all radicals.

1. 16 and 32
2. 45 and 35
3. 10 and 14
4. 28 and 42
5. 40 and 100
6. 51 and 8

Find the length of the missing variable(s). Simplify all radicals.

1. Fill in the blanks of the proof for Theorem 8-5. Given: $\triangle ABD$ with $\overline{AC} \perp \overline {DB}$ and $\angle DAB$ is a right angle. Prove: $\triangle ABD \sim \triangle CBA \sim \triangle CAD$
Statement Reason
1. Given
2. $\angle DCA$ and $\angle ACB$ are right angles
3. $\angle DAB \cong \angle DCA \cong \angle ACB$
4. Reflexive PoC
5. AA Similarity Postulate
6. $B \cong \angle B$
7. $\triangle CBA \cong \triangle ABD$
8. $\triangle CAD \cong \triangle CBA$

1. $\frac{3}{x} = \frac{x}{27} \rightarrow x^2 = 81 \rightarrow x = 9$
2. $\frac{\sqrt{6}}{x} = \frac{x}{9 \sqrt{6}} \rightarrow x^2 = 54 \rightarrow x = \sqrt{54} = \sqrt{9 \cdot 6} = 3 \sqrt{6}$
3. $\frac{x}{15} = \frac{12}{x} \rightarrow x^2 = 180 \rightarrow x = \sqrt{180} = \sqrt{4 \cdot 9 \cdot 5} = 2 \cdot 3 \sqrt{5} = 6 \sqrt{5}$
1. ${\;} \ 4^2+4^2 = h^2\!\\{\;} \ h = \sqrt{32} = 4 \sqrt{2}$

8 , 9 , 10

Feb 22, 2012

Aug 21, 2014