Question

Use the image to answer the 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=
Remaining Attempts : 1View previous attempt
Skip to navigation

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:

\[
\boxed{106}
\]