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) Responses x=8 x equals 8 x=3 x equals 3 x=4 x equals 4 x=6 x equals 6

Answer 1

In a parallelogram, the diagonals are congruent if it is a rectangle. Given that \( AC = 48 \) and \( DO = 7x + 3 \), we need to find the value of \( x \) such that:

\[
7x + 3 = 48
\]

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

\[
7x = 48 - 3
\]
\[
7x = 45
\]

Next, we divide both sides by 7:

\[
x = \frac{45}{7} \approx 6.43
\]

However, none of the choices you provided match that result exactly. Let's clarify how \( DO \) fits into the context of the diagonals. The notation \( DO \) should refer to one half of the diagonal \( AC \) if \( O \) is the midpoint. Since this could not be a standard setup because \( DO \) should be equal to half of \( AC \), re-examining the values gives:

Setting:

\[
DO = \frac{1}{2}AC = \frac{1}{2}(48) = 24
\]

We solve:

\[
7x + 3 = 24
\]

Subtract 3 again:

\[
7x = 21
\]

Then divide by 7:

\[
x = 3
\]

The value of \( x \) that proves parallelogram \( ABCD \) is a rectangle is:

**x equals 3**.