AEDC is similar to AABC. Solve for x. X =

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
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**Problem Statement:**

Given: \( \triangle EDC \) is similar to \( \triangle ABC \). Solve for \( x \).

**Solution:**

The key to solving this problem lies in understanding the property of similar triangles, where the corresponding sides have proportional lengths.

Given:
- \( \triangle EDC \) and \( \triangle ABC \) are similar.
- The sides of \( \triangle ABC \) are \( AB = 6 \, \text{m} \) and \( BC = 12.5 \, \text{m} \).
- The sides of \( \triangle EDC \) are \( ED = x \, \text{m} \) and \( DC = 4 \, \text{m} \).

**Steps to solve for \( x \):**

Since the triangles are similar:
\[ \frac{ED}{AB} = \frac{DC}{BC} \]

Plugging in the known values:
\[ \frac{x}{6} = \frac{4}{12.5} \]

Cross-multiplying to solve for \( x \):
\[ x = \frac{4 \cdot 6}{12.5} \]

\[ x = \frac{24}{12.5} \]

Simplifying:
\[ x = 1.92 \]

Therefore, the value of \( x \) is:
\[ x = 1.92 \, \text{m} \]

[ **Answer: \( x = 1.92 \, \text{m} \) **]

**Explanation of the Diagram:**

- The diagram depicts two right-angled triangles, \( \triangle ABC \) and \( \triangle EDC \).
- \( \triangle ABC \) is the larger triangle with a right angle at \( B \), and sides:
  - \( AB = 6 \, \text{m} \) (vertical side)
  - \( BC = 12.5 \, \text{m} \) (diagonal side)
- \( \triangle EDC \) is the smaller triangle with a right angle at \( D \), and sides:
  - \( ED = x \, \text{m} \) (vertical side)
  - \( DC = 4 \, \text{m} \) (horizontal side)

This detailed explanation provides the necessary steps to find the value of \( x \) using
Transcribed Image Text:**Problem Statement:** Given: \( \triangle EDC \) is similar to \( \triangle ABC \). Solve for \( x \). **Solution:** The key to solving this problem lies in understanding the property of similar triangles, where the corresponding sides have proportional lengths. Given: - \( \triangle EDC \) and \( \triangle ABC \) are similar. - The sides of \( \triangle ABC \) are \( AB = 6 \, \text{m} \) and \( BC = 12.5 \, \text{m} \). - The sides of \( \triangle EDC \) are \( ED = x \, \text{m} \) and \( DC = 4 \, \text{m} \). **Steps to solve for \( x \):** Since the triangles are similar: \[ \frac{ED}{AB} = \frac{DC}{BC} \] Plugging in the known values: \[ \frac{x}{6} = \frac{4}{12.5} \] Cross-multiplying to solve for \( x \): \[ x = \frac{4 \cdot 6}{12.5} \] \[ x = \frac{24}{12.5} \] Simplifying: \[ x = 1.92 \] Therefore, the value of \( x \) is: \[ x = 1.92 \, \text{m} \] [ **Answer: \( x = 1.92 \, \text{m} \) **] **Explanation of the Diagram:** - The diagram depicts two right-angled triangles, \( \triangle ABC \) and \( \triangle EDC \). - \( \triangle ABC \) is the larger triangle with a right angle at \( B \), and sides: - \( AB = 6 \, \text{m} \) (vertical side) - \( BC = 12.5 \, \text{m} \) (diagonal side) - \( \triangle EDC \) is the smaller triangle with a right angle at \( D \), and sides: - \( ED = x \, \text{m} \) (vertical side) - \( DC = 4 \, \text{m} \) (horizontal side) This detailed explanation provides the necessary steps to find the value of \( x \) using
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