A square with vertices upper A, upper S, upper D, and upper F (in that order) that appears slightly askew. There are two dashed diagonals that intersect at point upper W.

Question

If FW¯¯¯¯¯¯¯¯¯=13x−114 and SW¯¯¯¯¯¯¯¯¯=4x+21 , what value of AD¯¯¯¯¯¯¯¯ would be required in order for ASDF to be a square?

(1 point)
Responses

81
81

10.3¯¯¯
10 point repeating 3

15
15

162

Answer 1

In a square, the diagonals are equal in length. Given the lengths of the segments \( FW \) and \( SW \), we can set them equal to each other to find the value of \( x \):

\[ FW = SW \implies 13x - 114 = 4x + 21 \]

Now, we will solve for \( x \):

1. Move \( 4x \) to the left side:

\[ 13x - 4x - 114 = 21 \]

This simplifies to:

\[ 9x - 114 = 21 \]

2. Now, add \( 114 \) to both sides:

\[ 9x = 21 + 114 \]

Thus,

\[ 9x = 135 \]

3. Next, divide both sides by 9:

\[ x = \frac{135}{9} = 15 \]

Now that we have \( x \), we can substitute it back into either equation to find the length of one diagonal. Let’s use \( FW \):

\[ FW = 13x - 114 = 13(15) - 114 \]

Calculating \( FW \):

\[ FW = 195 - 114 = 81 \]

Since the diagonals of the square are equal, we then find:

\[ AD = FW = 81 \]

Thus, the value of \( AD \) required in order for \( ASDF \) to be a square is:

\[ \boxed{81} \]