Michael is 6 feet tall and is standing outside next to his younger sister. He notices that he can see both of their shadows and decides to measure each shadow. His shadow is 8 feet long and his sister's shadow is 5 feet long. How tall is Michael's sister?

### Applications of Similar Triangles

If two triangles are similar, then their corresponding angles are congruent and their corresponding sides are proportional. There are three criteria for proving that triangles are similar:

**AA:**If two triangles have two pairs of congruent angles, then the triangles are similar.**SAS:**If two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar.**SSS:**If three sides of one triangle are proportional to three sides of another triangle, then the triangles are similar.

Once you know that two triangles are similar, you can use the fact that their corresponding sides are proportional and their corresponding angles are congruent to solve problems.

Let's take a look at some example problems.

1. Prove that the two triangles below are similar.

The triangles are similar by \begin{align*}AA \sim\end{align*} because they have at least two pairs of congruent angles.

Use the Pythagorean Theorem to find \begin{align*}DE\end{align*}.

\begin{align*}\left (3 \sqrt{3} \right )^2 + DE^2 &= 6^2 \\ \rightarrow 27+DE^2 &= 36 \\ \rightarrow DE^2 &= 9 \\ \rightarrow DE &= 3 \end{align*}

Use the fact that the triangles are similar to find the missing sides of \begin{align*}\Delta ABC\end{align*}.

\begin{align*}\frac{AC}{DF} = \frac{18}{6} = 3\end{align*}, so the scale factor is 3.

- \begin{align*}\frac{AB}{DE} = 3 \rightarrow \frac{AB}{3} = 3 \rightarrow AB = 9\end{align*}
- \begin{align*}\frac{BC}{EF} = 3 \rightarrow\frac{BC}{3 \sqrt{3}} = 3 \rightarrow BC = 9 \sqrt{3}\end{align*}

#### The triangles are called 30-60-90 triangles because of their angles measures.

Explain why all 30-60-90 triangles are similar.

All 30-60-90 triangles are similar by \begin{align*}AA \sim\end{align*} because they will all have at least two pairs of congruent angles.

2. Use the triangles from #1 to find the ratios between the three sides of any 30-60-90 triangle.

\begin{align*}\Delta DEF\end{align*} had sides \begin{align*} 3,3 \sqrt{3},6\end{align*}. This ratio of \begin{align*}3:3 \sqrt{3}:6\end{align*} reduces to \begin{align*}1:\sqrt{3}:2\end{align*}. The three sides of any 30-60-90 triangle will be in this ratio.

3. Find the missing sides of the triangle below.

The side opposite the \begin{align*}30^\circ\end{align*} angle is the smallest side because \begin{align*}30^\circ\end{align*} is the smallest angle. Therefore, the length of 10 corresponds to the length of “1” in the ratio \begin{align*}1:\sqrt{3}:2\end{align*}. The scale factor is 10. The other sides of the triangle will be \begin{align*}10 \sqrt{3 }\end{align*} and 20, because \begin{align*}10:10 \sqrt{3}:20\end{align*} is equivalent to \begin{align*}1: \sqrt{3}:2\end{align*}. \begin{align*}BC=10 \sqrt{3}\end{align*} and \begin{align*}AC=20\end{align*}.

4. Create similar triangles in order to solve for \begin{align*}x\end{align*}.

Extend \begin{align*}\overline{AD}\end{align*} and \begin{align*}\overline{BC}\end{align*} to create point \begin{align*}G\end{align*}.

\begin{align*}\Delta DGC \sim \Delta EGF \sim \Delta AGB\end{align*} by \begin{align*}AA \sim\end{align*} because angles \begin{align*}\angle DCG, \angle EFG, \angle ABG\end{align*} are all right angles and are therefore congruent and all triangles share \begin{align*}\angle G\end{align*}. This means that their corresponding sides are proportional. First, solve for \begin{align*}GC\end{align*} by looking at \begin{align*}\Delta DGC\end{align*} and \begin{align*}\Delta EGF\end{align*}.

\begin{align*}\frac{2}{GC} &= \frac{3.5}{2.5 + GC} \\ 5 + 2GC &= 3.5 GC \\ 5 &= 1.5 GC \\ GC & \approx \ 3.33 \end{align*}

Next, solve for \begin{align*}x\end{align*} by looking at \begin{align*}\Delta DGC\end{align*} and \begin{align*}\Delta AGB\end{align*}.

\begin{align*}\frac{2}{3.33} &= \frac{x}{1.5+2.5+3.33} \\ \frac{2}{3.33} &= \frac{x}{7.33} \\ x &= 4.4 \end{align*}

**Examples**

**Example 1**

Earlier, you were asked how tall Michael's sister is. You can answer this question using applications of similar triangles.

The sun creates shadows at the same angle for both Michael and his sister. Assuming they are both standing up straight and making right angles with the ground, similar triangles are created.

Corresponding sides are proportional because the triangles are similar.

\begin{align*}\frac{\text{Michael's Height}}{\text{Sister's Height}} &= \frac{\text{Length of Michael's Shadow}}{\text{Length of Sister's Shadow}} \\ \frac{6 \ ft}{\text{Sister's Height}} &= \frac{8 \ ft}{5 \ ft} \end{align*}

Cross multiply and solve for his sister's height. His sister is 3.75 feet tall.

#### Example 2

Prove that all isosceles right triangles are similar.

Consider two generic isosceles right triangles:

Two pairs of sides are proportional with a ratio of \begin{align*}\frac{b}{a}\end{align*}. Also, \begin{align*}\angle C \cong \angle F\end{align*}. Therefore, the two triangles are similar by \begin{align*}SAS \sim\end{align*}.

#### Example 3

Find the measures of the angles of an isosceles right triangle. Why are isosceles right triangles called 45-45-90 triangles?

The base angles of an isosceles triangle are congruent. If the vertex angle is \begin{align*}90^\circ \end{align*}, each base angle is \begin{align*}\frac{180^\circ - 90^\circ}{2} = 45^\circ\end{align*}. The measures of the angles of an isosceles right triangle are 45, 45, and 90. An isosceles right triangle is called a 45-45-90 triangle because those are its angle measures.

#### Example 4

Use the Pythagorean Theorem to find the missing side of an isosceles right triangle whose legs are each length \begin{align*}x\end{align*}.

The missing side is the hypotenuse of the right triangle, \begin{align*}c\end{align*}. By the Pythagorean Theorem, \begin{align*}c^2 = x^2 + x^2\end{align*}. This means \begin{align*}c^2 = 2x^2 \end{align*} and therefore \begin{align*}c=x \sqrt{2}\end{align*}. The ratio of the sides of any isosceles right triangle will be \begin{align*}x:x:x \sqrt{2}\end{align*} which simplifies to \begin{align*} 1:1:\sqrt{2}\end{align*}.

#### Example 5

Use what you have learned in #1-#3 to find the missing sides of the right triangle below without using the Pythagorean Theorem.

If one of the legs is 3, then the other leg is also 3, so \begin{align*}AC=3\end{align*}. The hypotenuse will be \begin{align*}3 \sqrt{2}\end{align*} following the pattern from #3, so \begin{align*}AB = 3 \sqrt{2}\end{align*}.

### Review

1. Explain why all 30-60-90 triangles are similar.

2. The ratio between the sides of any 30-60-90 triangle is ________:________:________.

Find the missing sides of each triangle:

3.

4.

5.

6. Explain why all 45-45-90 triangles are similar.

7. The ratio between the sides of any 45-45-90 triangle is ________:________:________.

Find the missing sides of each triangle:

8.

9.

10.

Use the figure below for #11 and #12.

11. Prove that \begin{align*}\Delta ABC \sim \Delta EFD\end{align*}.

12. Find the value of \begin{align*}x\end{align*}.

Use the figure below for #13-#15.

13. Find \begin{align*}m \angle ADB\end{align*}. What type of triangle is \begin{align*}\Delta ADB\end{align*}?

14. Find \begin{align*}BD\end{align*} and \begin{align*}AB\end{align*}.

15. Find \begin{align*}AC\end{align*}.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 6.7.