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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=
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Answer 1

In a rectangle, the diagonals bisect each other. This means that the lengths of segments \(AO\) and \(DO\) are equal because \(O\) is the midpoint of both diagonals \(AC\) and \(BD\).

Given:

• \(AO = 9x - 1\)
• \(DO = 6x + 17\)

Since \(AO = DO\), we can set up the equation:

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

Now, we'll solve for \(x\):

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

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

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

Now, we can substitute \(x = 6\) back into either expression for \(AO\) or \(DO\) to find the length of one of the segments. Let's use \(AO\):

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

Since \(DO\) must also equal this length, we verify:

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

Both segments \(AO\) and \(DO\) are indeed \(53\).

The length of diagonal \(AC\) is twice the length of segment \(AO\) (since \(O\) is the midpoint):

\[ AC = 2 \times AO = 2 \times 53 = 106 \]

Thus, the length of diagonal \(AC\) is:

\[ \boxed{106} \]