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
Question
![**The two triangles are similar.**
**What is the value of x?**
**Enter your answer in the box.**
\[ x = [ \_\_\_ ] \]
---
**Figure Explanation:**
The image displays two right-angled triangles that are similar to each other. The smaller triangle has one leg labeled as 3 units and the hypotenuse labeled as 12 units. The larger triangle has one leg labeled as \( 3x + 1 \) and the hypotenuse labeled as \( 4x \).
---
**Concept:**
To find the value of \( x \), we need to use the properties of similar triangles. In similar triangles, the ratios of the corresponding sides are equal.
For two similar triangles:
\[
\frac{\text{Side 1 of Triangle A}}{\text{Side 1 of Triangle B}} = \frac{\text{Side 2 of Triangle A}}{\text{Side 2 of Triangle B}} = \frac{\text{Side 3 of Triangle A}}{\text{Side 3 of Triangle B}}
\]
In this problem, we can equate the ratios of their corresponding sides to solve for \( x \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b717370-ac2a-4449-94d5-6954bef42bac%2F1605a1e3-b255-47db-b781-2abd862b5a5b%2Fhavmbjf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**The two triangles are similar.**
**What is the value of x?**
**Enter your answer in the box.**
\[ x = [ \_\_\_ ] \]
---
**Figure Explanation:**
The image displays two right-angled triangles that are similar to each other. The smaller triangle has one leg labeled as 3 units and the hypotenuse labeled as 12 units. The larger triangle has one leg labeled as \( 3x + 1 \) and the hypotenuse labeled as \( 4x \).
---
**Concept:**
To find the value of \( x \), we need to use the properties of similar triangles. In similar triangles, the ratios of the corresponding sides are equal.
For two similar triangles:
\[
\frac{\text{Side 1 of Triangle A}}{\text{Side 1 of Triangle B}} = \frac{\text{Side 2 of Triangle A}}{\text{Side 2 of Triangle B}} = \frac{\text{Side 3 of Triangle A}}{\text{Side 3 of Triangle B}}
\]
In this problem, we can equate the ratios of their corresponding sides to solve for \( x \).
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