Use the image to answer the question.

Question

A rectangle upper A upper B upper C upper D has diagonals between upper A upper C and upper B upper D. The diagonals insect at upper O, in the center. Segment upper A upper O is labeled 9 x minus 1. Segment upper D upper O is labeled 6 x plus 17.

Given the rectangle ABCD
as shown, find AC
if AO=9x−1
and DO=6x+17
.

(1 point)
AC=

Answer 1

To find the length of diagonal \( AC \) in rectangle \( ABCD \), we need to first calculate the values of segments \( AO \) and \( DO \).

Since \( O \) is the midpoint of the diagonals \( AC \) and \( BD \), we have \( AO = DO \).

From the problem, we know: \[ AO = 9x - 1 \] \[ DO = 6x + 17 \]

Setting the two expressions equal to each other gives us: \[ 9x - 1 = 6x + 17 \]

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

1. Subtract \( 6x \) from both sides: \[ 3x - 1 = 17 \]

2. Add \( 1 \) to both sides: \[ 3x = 18 \]

3. Divide by \( 3 \): \[ x = 6 \]

Now that we have \( x \), we can substitute back into either equation to find \( AO \) or \( DO \). Let's find \( AO \): \[ AO = 9x - 1 = 9(6) - 1 = 54 - 1 = 53 \]

Since \( AC \) is made up of both segments \( AO \) and \( OC \) (where \( OC \) is equal to \( AO \) as they are both half of the diagonal), the diagonal \( AC \) is: \[ AC = AO + OC = AO + AO = 2 \times AO = 2 \times 53 = 106 \]

Thus, the length of diagonal \( AC \) is: \[ \boxed{106} \]