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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

x=8
x equals 8

x=4
x equals 4

x=6
x equals 6

x=3
x equals 3
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1 answer

To determine the value of \( x \) that proves parallelogram ABCD is a rectangle given that \( AC = 48 \) and \( DO = 7x + 3 \), we need to use the property that the diagonals of a rectangle are congruent (they have the same length).

From the context, it seems that \( DO \) is half the length of diagonal \( AC \) in parallelogram ABCD since \( O \) is the midpoint of diagonal \( AC \). Therefore, we have:

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

Now we can set up the equation:

\[ 7x + 3 = 24 \]

Next, we solve for \( x \):

  1. Subtract 3 from both sides:

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

  1. Divide both sides by 7:

\[ x = \frac{21}{7} = 3 \]

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

x = 3