---## Solving for $ x $ in Similar TrianglesIn this example, we are given two triangles that are similar, and we need to solve for the variable $ x $.### Problem Statement:**Given:**Triangles $\triangle ABC \sim \triangle UTS$**Details:**- In $\triangle UTS$: - $ UT = 16 $ - $ ST = x + 11 $ - $ SU $ is not labeled- In $\triangle ABC$: - $ AC = 40 $ - $ BC = 55 $ - $ AB $ is not labeled**With Options:**- $A) \; 5$- $B) \; 4$- $C) \; 7$- $D) \; 11$### Diagrams Explanation:The problem includes diagrams of two triangles with specific side lengths labeled:1. **$\triangle UTS$**: - Side $UT$ is given as 16 units. - Side $ST$ is expressed as $x + 11$.2. **$\triangle ABC$**: - Side $AC$ measures 40 units. - Side $BC$ measures 55 units.### Solution Approach:Since the two triangles are similar, the corresponding sides are proportional. This means for corresponding sides:\[ \frac{UT}{AC} = \frac{ST}{BC} \]Substitute the known values:\[ \frac{16}{40} = \frac{x + 11}{55} \]Solve the proportion for $ x $:1. Simplify the left side:\[ \frac{16}{40} = \frac{2}{5} \]2. Set up the equation:\[ \frac{2}{5} = \frac{x + 11}{55} \]3. Cross multiply:\[ 2 \times 55 = 5 \times (x + 11) \]\[ 110 = 5x + 55 \]4. Solve for $ x $:\[ 110 - 55 = 5x \]\[ 55 = 5x \]\[ x = \frac{55}{5} \]\[ x = 11 \]So, the correct answer is:**\(

Answered: Solve for x. The triangles in each pair…

Math

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Solve for x. The triangles in each pair are similar. 6) AABC AUTS U 16 x+11 40 A B 55 В) 4 D) 11 A) 5 C) 7

Elementary Geometry For College Students, 7e

7th Edition

ISBN:9781337614085

Author:Alexander, Daniel C.; Koeberlein, Geralyn M.

Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.

ChapterP: Preliminary Concepts

SectionP.CT: Test

Problem 1CT

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Question

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## Solving for $ x $ in Similar Triangles

In this example, we are given two triangles that are similar, and we need to solve for the variable $ x $.

### Problem Statement:

**Given:**
Triangles $\triangle ABC \sim \triangle UTS$

**Details:**

- In $\triangle UTS$:
- $ UT = 16 $
- $ ST = x + 11 $
- $ SU $ is not labeled

- In $\triangle ABC$:
- $ AC = 40 $
- $ BC = 55 $
- $ AB $ is not labeled

**With Options:**
- $A) \; 5$
- $B) \; 4$
- $C) \; 7$
- $D) \; 11$

### Diagrams Explanation:

The problem includes diagrams of two triangles with specific side lengths labeled:

1. **$\triangle UTS$**:
- Side $UT$ is given as 16 units.
- Side $ST$ is expressed as $x + 11$.

2. **$\triangle ABC$**:
- Side $AC$ measures 40 units.
- Side $BC$ measures 55 units.

### Solution Approach:

Since the two triangles are similar, the corresponding sides are proportional. This means for corresponding sides:

\[ \frac{UT}{AC} = \frac{ST}{BC} \]

Substitute the known values:

\[ \frac{16}{40} = \frac{x + 11}{55} \]

Solve the proportion for $ x $:

1. Simplify the left side:
\[ \frac{16}{40} = \frac{2}{5} \]

2. Set up the equation:
\[ \frac{2}{5} = \frac{x + 11}{55} \]

3. Cross multiply:
\[ 2 \times 55 = 5 \times (x + 11) \]
\[ 110 = 5x + 55 \]

4. Solve for $ x $:
\[ 110 - 55 = 5x \]
\[ 55 = 5x \]
\[ x = \frac{55}{5} \]
\[ x = 11 \]

So, the correct answer is:

**\(

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