# Application of Similar Triangles

 Site: Saylor Academy Course: GKT101: General Knowledge for Teachers – Math Book: Application of Similar Triangles
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## Description

Read this section and watch the videos to see the examples of applications of similar triangles in geometric and real-life problems.

## Applications of Similar Triangles

Similar triangles have congruent corresponding angles and proportional corresponding sides.
There are three common sets of criteria for proving that triangles are similar:
1. AA: If two triangles have two pairs of congruent angles, then the triangles are similar.

[Figure 1]

2. 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.

[Figure 2]

3. SSS: If three sides of one triangle are proportional to three sides of another triangle, then the triangles are similar.

[Figure 3]

## Triangle Similarity and 30-60-90 Triangles

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.

Consider the triangles in the image below.

a. Prove that the triangles are similar.

The triangles are similar by $A A \sim$ because they have at least two pairs of congruent angles, i.e.,

$\angle A B C=\angle D E F=90^{\circ}, \angle B C A=\angle E F D=30^{\circ}, \angle C A B=\angle F D E=60^{\circ}$

b. Use the Pythagorean Theorem to find $D E$.

According to the Pythagorean Theorem, in $\triangle D E F$ :

\begin{aligned}E F^{2}+D E^{2} &=F D^{2} \\(3 \sqrt{3})^{2}+D E^{2} &=6^{2} \\(9 \cdot 3)+D E^{2} &=6^{2} \\27+D E^{2} &=36 \\D E^{2} &=36-27 \\D E^{2} &=9 \\D E &=3\end{aligned}

c. Use the fact that the triangles are similar to find the missing sides of $\triangle A B C$.

$\triangle A B C \sim \triangle D E F$, which means that the corresponding sides are proportional. Therefore,

\begin{aligned}&\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F} \\&\frac{A C}{D F}=\frac{18}{6}=3, \text { so the scale factor is } 3\end{aligned}

• $\frac{A B}{D E}=3 \rightarrow \frac{A B}{3}=3 \rightarrow A B=9$
• $\frac{B C}{E F}=3 \rightarrow \frac{B C}{3 \sqrt{3}}=3 \rightarrow B C=9 \sqrt{3}$

Triangles of this configuration are called $30-60-90$ triangles because of their angle measures.

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

All 30-60-90 triangles are similar by $A A \sim$ because they all have at least two pairs of congruent angles.

## 30-60-90 Triangle Side Ratios

1. Find the ratios between the three sides of any $30-60-90$ triangle.

[Figure 4]

From the reference image: $\triangle D E F$ has sides

$D E=3, E F=3 \sqrt{3}$, and $F D=6$. This ratio of $3: 3 \sqrt{3}: 6$ reduces to $1: \sqrt{3}: 2$.

The three sides of any 30-60-90 triangle will be in this ratio: $1: \sqrt{3}: 2$.

2. Find the missing sides of the triangle below.

[Figure 5]

The side opposite the $30^{\circ}$ angle is the smallest side because $30^{\circ}$ is the smallest angle. Therefore, the length of $10$ corresponds to the length of $1$ in the ratio $1: \sqrt{3}: 2$. The scale factor is $10$. The other sides of the triangle will be $10 \sqrt{3}$ and $20$, because $10: 10 \sqrt{3}: 20$ is equivalent to $1: \sqrt{3}: 2$. $B C=10 \sqrt{3}$ and $A C=20$.

## Similar Triangle Applications

Create similar triangles in order to solve for $x$.

[Figure 6]

Extend $\overline{A D}$ and $\overline{B C}$ to create point $G$.

[Figure 7]

$\triangle D G C \sim \triangle E G F \sim \triangle A G B$ by $A A \sim$ because angles $\angle D C G, \angle E F G, \angle A B G$ are all right angles and are therefore congruent and all triangles share $\angle G$. This means that their corresponding sides are proportional. First, solve for $G C$ by looking at $\triangle D G C$ and $\triangle E G F . \triangle D G C \sim \triangle E G F$ which means that corresponding sides are proportional.

Therefore:

\begin{aligned} \frac{D G}{E G} &=\frac{G C}{G F}=\frac{D C}{E F} \\ \frac{D C}{G C} &=\frac{E F}{G F} \\ \frac{2}{G C} &=\frac{3.5}{2.5+G C} \\ 5+2 G C &=3.5 G C \\ 5 &=1.5 G C \\ G C & \approx 3.33 \end{aligned}

Next, solve for $x$ by looking at $\triangle D G C$ and $\triangle A G B . \triangle D G C \sim \triangle A G B$ which means that corresponding sides are proportional.

Therefore:

\begin{aligned} \frac{D G}{A G} &=\frac{G C}{G B}=\frac{D C}{A B} \\ \frac{D C}{G C} &=\frac{A B}{G B} \\ \frac{2}{3.33} &=\frac{x}{1.5+2.5+3.33} \\ \frac{2}{3.33} &=\frac{x}{7.33} \\ x &=4.4 \end{aligned}

## Examples

#### Example 1

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?

You can answer this question by applying what you know about 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.

[Figure 7]

Corresponding sides are proportional because the triangles are similar.

\begin{aligned}\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 } \\\text { Sister's Height } &=\frac{6 \cdot 5}{8} \\&=3.75 ft\end{aligned}

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 $\frac{b}{a}$. Also, $\angle C \cong \angle F$. Therefore, the two triangles are similar by $S A S \sim$.

#### 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 $90^{\circ}$, each base angle is $\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}$. 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 $x$.

[Figure 9]

The missing side is the hypotenuse of the right triangle, labeled $c$.

By the Pythagorean Theorem,

\begin{aligned}A B^{2} &=A C^{2}+B C^{2} \\c^{2} &=x^{2}+x^{2} \\c^{2} &=2 x^{2} \\c &=\sqrt{2 x^{2}} \\c &=x \sqrt{2}\end{aligned}

• The ratio of the sides of any isosceles right triangle will be $x: x: x \sqrt{2}$ which simplifies to $1: 1: \sqrt{2}$.

#### Example 5

Find the missing sides of the right triangle below without using the Pythagorean Theorem.

[Figure 10]

If one of the legs is $3$, then the other leg is also $3$, so $A C=3$. The ratio of the sides of an isosceles right triangle is $1: 1: \sqrt{2}$. Therefore, the hypotenuse will be $3 \sqrt{2}$, so $A B=3 \sqrt{2}$.