Question

Because congruent diagonals prove that a parallelogram is a rectangle, what value of x proves that parallelogram ABCD is a rectangle if AC=48 and DO=7x+3
(1 point)
x = 8
x = 4
x = 3
x = 6

Answer 1

In a parallelogram, the diagonals bisect each other. This means that the lengths of the segments created by the intersection of the diagonals are equal.

Given that diagonal AC = 48, the length of each segment from the intersection point O to the endpoints of the diagonal AC (which is bisected at point O) is:

\[
AO = OC = \frac{AC}{2} = \frac{48}{2} = 24
\]

We also have that \(DO = 7x + 3\). Since AO and DO are equal in a parallelogram with congruent diagonals, we can set up the equation:

\[
7x + 3 = 24
\]

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

1. Subtract 3 from both sides:

\[
7x = 24 - 3
\]
\[
7x = 21
\]

2. Divide both sides by 7:

\[
x = \frac{21}{7}
\]
\[
x = 3
\]

Therefore, the value of \(x\) that proves that parallelogram ABCD is a rectangle is:

\[
\boxed{3}
\]