A parallelogram with vertices upper A upper B upper C upper D.

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=6
x equals 6

x=4
x equals 4

x=3

Answer 1

To find the value of \( x \) that proves that parallelogram ABCD is a rectangle based on the given information about the diagonals, we need to set the lengths of the diagonals equal to each other.

In a parallelogram, the diagonals bisect each other, so let \( O \) be the midpoint of both diagonals \( AC \) and \( BD \). Given that \( AC = 48 \), the length of each half of diagonal \( AC \) is:

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

The diagonal \( BD \) has a length represented as \( DO \), which is given as \( DO = 7x + 3 \). Since \( O \) is the midpoint, we also have:

\[ BO = DO \]

Thus, the entire diagonal \( BD \) is:

\[ BD = BO + DO = DO + DO = 2 \cdot DO = 2(7x + 3) = 14x + 6 \]

For the parallelogram ABCD to be a rectangle, the diagonals must be equal:

\[ AC = BD \]

So we have the equation:

\[ 48 = 14x + 6 \]

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

1. Subtract 6 from both sides: \[ 48 - 6 = 14x \] \[ 42 = 14x \]

2. Now, divide both sides by 14: \[ x = \frac{42}{14} = 3 \]

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

\(\boxed{3}\)