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 = 6 x = 4 x = 8 x = 3

Explanation:

Step1: Properties of rectangle diagonals

In a rectangle, diagonals are congruent and bisect each other. So \(AC = BD\) and \(DO=\frac{1}{2}BD\). Since \(AC = 48\), then \(BD=48\) and \(DO=\frac{1}{2}\times48 = 24\).

Step2: Solve for \(x\)

Given \(DO = 7x+3\), and \(DO = 24\). Set up the equation \(7x+3=24\). Subtract 3 from both sides: \(7x=24 - 3=21\). Divide both sides by 7: \(x=\frac{21}{7}=3\).

Answer:

\(x = 3\)