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
Related questions
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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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Transcribed Image Text:---
## 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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