Question

There is a parallelogram with 5 points, top left is A, top right is D, bottom left is B, bottom right is C, and O is in the middle of 2 diagonal lines inside the parallelogram

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

x=4
x=8
x=6
x=3

Answer 1

In a parallelogram, the diagonals bisect each other. Therefore, in parallelogram ABCD, point O, being the intersection of the diagonals AC and BD, indicates that AO = OC and BO = OD.

Since you mentioned that \( AC = 48 \), point O being the midpoint means that:

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

Now we also know that \( DO = 7x + 3 \) and using the fact that diagonals bisect each other, we have:

\[
BO = OD
\]

Because AO and DO refer to the same segments from O to the corners, we apply the equality:

\[
AO = DO
\]
\[
24 = 7x + 3
\]

To solve for \( x \), we will subtract 3 from both sides:

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

Next, divide both sides by 7:

\[
x = 3
\]

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

\[
\boxed{3}
\]