ABCD is a square. Find the value of x. A D (11x + 35) In a square, the diagonals are X =

Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
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ABCD is a square. Find the value of x.

### Problem Description

ABCD is a square. Find the value of \( x \).

### Diagram Explanation

The diagram represents a square ABCD with lines drawn from corner to corner, forming two diagonals. These diagonals intersect at the center, creating four right-angle triangles within the square. At the intersection point of the diagonals, an angle is marked as \((11x + 35)^\circ\).

### Key Points
- In a square, the diagonals are **perpendicular bisectors** of each other.
- This means they form four right angles (90 degrees) at the intersection.

### Problem Solution

Given that the angle at the intersection of the diagonals is \( (11x + 35)^\circ \):

Since the diagonals are perpendicular, they create a 90-degree angle.

Set the expression equal to 90 degrees:
\[ 11x + 35 = 90 \]

Solve for \( x \):

\[ 11x = 90 - 35 \]  
\[ 11x = 55 \]  
\[ x = \frac{55}{11} \]  
\[ x = 5 \]

Thus, the value of \( x \) is:

\[ x = 5 \]
Transcribed Image Text:### Problem Description ABCD is a square. Find the value of \( x \). ### Diagram Explanation The diagram represents a square ABCD with lines drawn from corner to corner, forming two diagonals. These diagonals intersect at the center, creating four right-angle triangles within the square. At the intersection point of the diagonals, an angle is marked as \((11x + 35)^\circ\). ### Key Points - In a square, the diagonals are **perpendicular bisectors** of each other. - This means they form four right angles (90 degrees) at the intersection. ### Problem Solution Given that the angle at the intersection of the diagonals is \( (11x + 35)^\circ \): Since the diagonals are perpendicular, they create a 90-degree angle. Set the expression equal to 90 degrees: \[ 11x + 35 = 90 \] Solve for \( x \): \[ 11x = 90 - 35 \] \[ 11x = 55 \] \[ x = \frac{55}{11} \] \[ x = 5 \] Thus, the value of \( x \) is: \[ x = 5 \]
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