Question

use the image to answer the 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)

x = 3
x = 4
x = 6
x = 8

Explanation:

Step1: Recall the property of rectangle diagonals

In a rectangle, the diagonals are congruent and bisect each other. So \(AC = BD\) and \(DO=\frac{1}{2}BD\), \(AC = 48\), then \(DO=\frac{1}{2}AC\).

Step2: Set up the equation

Since \(DO=\frac{1}{2}AC\) and \(AC = 48\), we have \(DO = 24\). Given \(DO=7x + 3\), then the equation is \(7x+3=24\).

Step3: Solve the equation for \(x\)

Subtract 3 from both sides of the equation: \(7x=24 - 3\), so \(7x=21\). Divide both sides by 7: \(x=\frac{21}{7}=3\).

Answer:

\(x = 3\) (the first option)