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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 the diagonal AC in rectangle ABCD, we first need to find the value of \(x\) by setting the lengths AO and DO equal to each other, since diagonals in a rectangle bisect each other.

We have:

• \(AO = 9x - 1\)
• \(DO = 6x + 17\)

Setting them equal to each other:

\[ 9x - 1 = 6x + 17 \]

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

1. Subtract \(6x\) from both sides: \[ 9x - 6x - 1 = 17 \] \[ 3x - 1 = 17 \]

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

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

Now that we have \(x\), we can substitute it back into either expression to find AO and DO:

Calculating AO: \[ AO = 9x - 1 = 9(6) - 1 = 54 - 1 = 53 \]

Calculating DO: \[ DO = 6x + 17 = 6(6) + 17 = 36 + 17 = 53 \]

Both segments AO and DO are confirmed to be equal to \(53\).

To find the length of diagonal AC, we can use the relationship: \[ AC = AO + OC = AO + AO = 2 \cdot AO = 2 \cdot 53 = 106 \]

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