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?

Answer 1

In any parallelogram, the diagonals bisect each other. For parallelogram ABCD, let's denote the intersection point of the diagonals AC and BD as point O. Since O is the midpoint of both diagonals, we have:

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

Given that segment DO is expressed as \( DO = 7x + 3 \), and since O is the midpoint of diagonal BD, we can set the length of segment DO to be half of diagonal BD. Therefore, we have:

\[
DO = \frac{BD}{2}
\]

For the diagonals to be congruent, we need \( AO = DO \). Since we already calculated \( AO \) as 24, we can set up the equation:

\[
7x + 3 = 24
\]

Now, let's solve for \( x \):

1. Subtract 3 from both sides:
\[
7x = 24 - 3
\]
\[
7x = 21
\]

2. Now, divide both sides by 7:
\[
x = \frac{21}{7}
\]
\[
x = 3
\]

Thus, the value of \( x \) that proves that parallelogram ABCD is a rectangle is \( \boxed{3} \).