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# 8.3: Using Similar Right Triangles

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify similar triangles inscribed in a larger triangle.
• Evaluate the geometric mean.
• Find the length of an altitude or leg using the geometric mean.

## Review Queue

1. If two triangles are right triangles, does that mean they are similar? Explain.
2. If two triangles are isosceles right triangles, does that mean they are similar? Explain.
3. Solve the ratio: \begin{align*}\frac{3}{x} = \frac{x}{27}\end{align*}.
4. 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? In California, the average home price increased 21.3% in 2004 and another 16.0% in 2005. What is the average rate of increase for these two years?

## Inscribed Similar Triangles

You may recall that if two objects are similar, corresponding angles are congruent and their sides are proportional in length. Let’s look at a right triangle, with an altitude drawn from the right angle.

There are three right triangles in this picture, \begin{align*}\triangle ADB, \triangle CDA\end{align*}, and \begin{align*}\triangle CAB\end{align*}. Both of the two smaller triangles are similar to the larger triangle because they each share an angle with \begin{align*}\triangle ADB\end{align*}. That means all three triangles are similar to each other.

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.

The proof of Theorem 8-5 is in the review questions.

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

Solution: If \begin{align*}m \angle E = 30^\circ\end{align*}, then \begin{align*}m \angle I = 60^\circ\end{align*} and \begin{align*}m \angle TRE = 60^\circ\end{align*}. \begin{align*}m \angle IRT = 30^\circ\end{align*} because it is complementary to \begin{align*}\angle TRE\end{align*}. Line up the congruent angles in the similarity statement. \begin{align*}\triangle IRE \sim \triangle ITR \sim \triangle RTE\end{align*}

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

Example 2: Find the value of \begin{align*}x\end{align*}.

Solution: First, let’s separate the triangles to find the corresponding sides.

Now we can set up a proportion.

\begin{align*}\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\end{align*}

Example 3: Find the value of \begin{align*}x\end{align*}.

Solution: Let’s set up a proportion.

\begin{align*}\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}\end{align*}

Example 4: Find the value of \begin{align*}y\end{align*} in \begin{align*}\triangle RST\end{align*} above.

Solution: Use the Pythagorean Theorem.

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

## The Geometric Mean

You are probably familiar with the arithmetic mean, which divides the sum of \begin{align*}n\end{align*} numbers by \begin{align*}n\end{align*}. This is commonly used to determine the average test score for a group of students.

The geometric mean is a different sort of average, which takes the \begin{align*}n^{th}\end{align*} root of the product of \begin{align*}n\end{align*} numbers. In this text, we will primarily compare two numbers, so we would be taking the square root of the product of two numbers. This mean is commonly used with rates of increase or decrease.

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

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

Solution: \begin{align*}x = \sqrt{24 \cdot 36} = \sqrt{12 \cdot 2 \cdot 12 \cdot 3} = 12 \sqrt{6}\end{align*}

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

Solution: \begin{align*}x = \sqrt{18 \cdot 54} = \sqrt{18 \cdot 18 \cdot 3} = 18 \sqrt{3}\end{align*}

Notice that in both of these examples, we did not actually multiply the two numbers together, but kept them separate. This made 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 \begin{align*}x\end{align*}.

Solution: Using similar triangles, we have the proportion

\begin{align*}\frac{\text{shortest leg of smallest}\ \triangle}{\text{shortest leg of middle}\ \triangle} & = \frac{\text{longer leg of smallest}\ \triangle}{\text{longer leg of middle}\ \triangle}\\ \frac{9}{x} & = \frac{x}{27}\\ x^2 & = 243\\ x & = \sqrt{243} = 9 \sqrt{3}\end{align*}

In Example 7, \begin{align*}\frac{9}{x} = \frac{x}{27}\end{align*} 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.

Theorem 8-6: In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of the altitude is the geometric mean of these two segments.

Theorem 8-7: In a right triangle, the altitude drawn from the right angle to the hypotenuse divides the hypotenuse into two segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

In other words

Theorem 8-6: \begin{align*}\frac{BC}{AC} = \frac{AC}{DC}\end{align*}

Theorem 8-7: \begin{align*}\frac{BC}{AB} = \frac{AB}{DB}\end{align*} and \begin{align*}\frac{DC}{AD} = \frac{AD}{DB}\end{align*}

Both of these theorems are proved using similar triangles.

Example 8: Find the value of \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Solution: Use theorem 8-7 to solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

\begin{align*}\frac{20}{x} &= \frac{x}{35} && \ \frac{15}{y} = \frac{y}{35}\\ x^2 &= 20 \cdot 35 && \ \ y^2 = 15 \cdot 35\\ x &= \sqrt{20 \cdot 35} && \quad y = \sqrt{15 \cdot 35}\\ x &= 10 \sqrt{7} && \quad y = 5 \sqrt{21}\end{align*}

You could also use the Pythagorean Theorem to solve for \begin{align*}y\end{align*}, once \begin{align*}x\end{align*} has been solved for.

\begin{align*}\left ( 10 \sqrt{7} \right )^2 + y^2 & = 35^2\\ 700 + y^2 & = 1225\\ y & = \sqrt{525} = 5 \sqrt{21}\end{align*}

Either method is acceptable.

Know What? Revisited The average rate of increase can be found by using the geometric mean.

\begin{align*}x = \sqrt{0.213 \cdot 0.16} = 0.1846\end{align*}

Over the two year period, housing prices increased 18.46%.

## Review Questions

Use the diagram to answer questions 1-4.

1. Write the similarity statement for the three triangles in the diagram.
2. If \begin{align*}JM = 12\end{align*} and \begin{align*}ML = 9\end{align*}, find \begin{align*}KM\end{align*}.
3. Find \begin{align*}JK\end{align*}.
4. Find \begin{align*}KL\end{align*}.

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. Write a proof for Theorem 8-5. Given: \begin{align*}\triangle ABD\end{align*} with \begin{align*}\overline{AC} \bot \overline{DB}\end{align*} and \begin{align*}\angle DAB\end{align*} is a right angle. Prove: \begin{align*}\triangle ABD \sim \triangle CBA \sim \triangle CAD\end{align*}
2. Fill in the blanks for the proof of Theorem 8-7. Given: \begin{align*}\triangle ABD\end{align*} with \begin{align*}\overline{AC} \bot \overline{DB}\end{align*} and \begin{align*}\angle DAB\end{align*} is a right angle. Prove: \begin{align*}\frac {BC}{AB}=\frac {AB}{DB}\end{align*}
Statement Reason
1. \begin{align*}\triangle ABD\end{align*} with \begin{align*}\overline{AC} \bot \overline{DB}\end{align*} and \begin{align*}\angle DAB\end{align*} is a right angle.
2. \begin{align*}\triangle ABD \sim \triangle CBA \sim \triangle CAD\end{align*}
3. \begin{align*}\frac{BC}{AB} = \frac{AB}{DB}\end{align*}
1. Last year Poorva’s rent increased by 5% and this year her landlord wanted to raise her rent by 7.5%. What is the average rate at which her landlord has raised her rent over the course of these two years?
2. Mrs. Smith teaches AP Calculus. Between the first and second years she taught the course her students’ average score improved by 12%. Between the second and third years, the scores increased by 9%. What is the average rate of improvement in her students’ scores?
3. According to the US Census Bureau, http://www.census.gov/ipc/www/idb/country.php the rate of growth of the US population was 0.8% and in 2009 it was 1.0%. What was the average rate of population growth during that time period?

Algebra Connection A geometric sequence is a sequence of numbers in which each successive term is determined by multiplying the previous term by the common ratio. An example is the sequence 1, 3, 9, 27, ... Here each term is multiplied by 3 to get the next term in the sequence. Another way to look at this sequence is to compare the ratios of the consecutive terms.

1. Find the ratio of the \begin{align*}2^{nd}\end{align*} to \begin{align*}1^{st}\end{align*} terms and the ratio of the \begin{align*}3^{rd}\end{align*} to \begin{align*}2^{nd}\end{align*} terms. What do you notice? Is this true for the next set (\begin{align*}4^{th}\end{align*} to \begin{align*}3^{rd}\end{align*} terms)?
2. Given the sequence 4, 8, 16,..., if we equate the ratios of the consecutive terms we get: \begin{align*}\frac{8}{4} = \frac{16}{8}\end{align*}. This means that 8 is the ___________________ of 4 and 16. We can generalize this to say that every term in a geometric sequence is the ___________________ of the previous and subsequent terms.

Use what you discovered in problem 26 to find the middle term in the following geometric sequences.

1. 5, ____, 20
2. 4, ____, 100
3. 2, ____, \begin{align*}\frac{1}{2}\end{align*}
4. We can use what we have learned in this section in another proof of the Pythagorean Theorem. Use the diagram to fill in the blanks in the proof below.
Statement Reason
1. \begin{align*}\frac{e}{a} = \frac{?}{d + e}\end{align*} and \begin{align*}\frac{d}{b} = \frac{b}{?}\end{align*} Theorem 8-7
2. \begin{align*}a^2 = e(d + e)\end{align*} and \begin{align*}b^2 = d(d + e)\end{align*} \begin{align*}?\end{align*}
3. \begin{align*}a^2 + b^2 = ?\end{align*} Combine equations from #2.
4. \begin{align*}?\end{align*} Distributive Property
5. \begin{align*}c = d + e\end{align*} \begin{align*}?\end{align*}
6. \begin{align*}?\end{align*} Substitution PoE

1. No, another angle besides the right angles must also be congruent.
2. Yes, the three angles in an isosceles right triangle are \begin{align*}45^\circ, 45^\circ,\end{align*} and \begin{align*}90^\circ\end{align*}. Isosceles right triangles will always be similar.
3. \begin{align*}\frac{3}{x} = \frac{x}{27} \rightarrow x^2 = 81 \rightarrow x = \pm 9\end{align*}
4. \begin{align*}4^2 + 4^2 = h^2\!\\ h = \sqrt{32} = 4 \sqrt{2}\end{align*}

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Date Created:
Feb 22, 2012