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

AcellusIn order for the parallelogram to be arectangle, x = [?].Diagonal AC 11x + 8Diagonal-example-1
User Prabh Deep
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Solution

In order for a parallelogram to be a rectangle, the diagonals should have the same length


\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}

User Jacob Dalton
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