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?
X=4
X=3
X=6
X=8
X=4
X=3
X=6
X=8
Answers
GPT-4o mini
In a parallelogram, the diagonals bisect each other. This means that if you have diagonals AC and BD intersecting at point O, then:
\( AO = OC \) and \( BO = OD \).
Given that \( AC = 48 \), we can find the lengths of \( AO \) and \( OC \):
\[
AO = OC = \frac{AC}{2} = \frac{48}{2} = 24.
\]
Now, we also know that \( DO = 7x + 3 \). Since \( O \) is the midpoint of diagonal \( BD \), it follows that \( DO = BO \). Hence, we can express \( BO \) in terms of \( DO \):
\[
BO = DO.
\]
Since triangles \( AOB \) and \( COD \) are congruent (due to the properties of a parallelogram), we can set \( BO \) equal to \( 24 \):
\[
7x + 3 = 24.
\]
Now, we solve for \( x \):
1. Subtract \( 3 \) from both sides:
\[
7x = 24 - 3 = 21.
\]
2. Divide both sides by \( 7 \):
\[
x = \frac{21}{7} = 3.
\]
Thus, the value of \( x \) that proves that parallelogram ABCD is a rectangle is:
\[
\boxed{3}.
\]
\( AO = OC \) and \( BO = OD \).
Given that \( AC = 48 \), we can find the lengths of \( AO \) and \( OC \):
\[
AO = OC = \frac{AC}{2} = \frac{48}{2} = 24.
\]
Now, we also know that \( DO = 7x + 3 \). Since \( O \) is the midpoint of diagonal \( BD \), it follows that \( DO = BO \). Hence, we can express \( BO \) in terms of \( DO \):
\[
BO = DO.
\]
Since triangles \( AOB \) and \( COD \) are congruent (due to the properties of a parallelogram), we can set \( BO \) equal to \( 24 \):
\[
7x + 3 = 24.
\]
Now, we solve for \( x \):
1. Subtract \( 3 \) from both sides:
\[
7x = 24 - 3 = 21.
\]
2. Divide both sides by \( 7 \):
\[
x = \frac{21}{7} = 3.
\]
Thus, the value of \( x \) that proves that parallelogram ABCD is a rectangle is:
\[
\boxed{3}.
\]