AcellusIn order for the parallelogram to be arectangle, x = [?].Diagonal AC 11x + 8Diagonal BD 5x + 62AB=DСEnter

Solution
In order for a parallelogram to be a rectangle, the diagonals should have the same length
[tex]\begin{gathered} 11x+8\text{ =5x+62} \\ \text{collect the }like\text{ terms} \\ 11x-5x=62-8 \\ 6x=54 \\ \text{Divide both sides by 6} \\ x=9 \end{gathered}[/tex]