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Question

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

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 how do you solve it

x=8
x equals 8

x=4
x equals 4

x=6
x equals 6

x=3
x equals 3
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Answer 1

To determine the value of \( x \) that proves the parallelogram \( ABCD \) is a rectangle by showing that its diagonals are congruent, we are given:

• \( AC = 48 \)
• \( DO = 7x + 3 \)

In a parallelogram, the diagonals bisect each other, which means that the length from one vertex to the midpoint is equal for both diagonals. Thus, \( DO \) should equal half of \( AC \).

First, we can find the length of half of diagonal \( AC \): \[ \frac{AC}{2} = \frac{48}{2} = 24 \]

Next, we set up the equation: \[ DO = 7x + 3 \] Setting this equal to \( 24 \): \[ 7x + 3 = 24 \]

Now, we solve for \( x \): Subtract \( 3 \) from both sides: \[ 7x = 24 - 3 \] \[ 7x = 21 \]

Now, divide by \( 7 \): \[ x = \frac{21}{7} = 3 \]

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