If the two diagonals of a parallelogram are 7 | vedclass

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Question

(A) Let the parallelogram be $ABCD$ with diagonals $AC = 72 \, cm$ and $BD = 30 \, cm$.In a parallelogram,the diagonals bisect each other at point $O$

Vedclass · Accepted Answer

$156$

Vedclass · Answer

(A) Let the parallelogram be $ABCD$ with diagonals $AC = 72 \, cm$ and $BD = 30 \, cm$.In a parallelogram,the diagonals bisect each other at point $O$.Therefore,$OA = OC = \frac{72}{2} = 36 \, cm$ and $OB = OD = \frac{30}{2} = 15 \, cm$.Since the diagonals of a rhombus bisect each other at right angles,if we assume this is a rhombus (or calculate the side length using the property of diagonals),the side length $s$ is given by $s = \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$.$s = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39 \, cm$.Since all four sides of a rhombus are equal,the perimeter is $4 	imes 39 = 156 \, cm$.Note: $A$ parallelogram with diagonals bisecting at right angles is a rhombus. If it is not a rhombus,the perimeter cannot be uniquely determined. Given the options,it is implied to be a rhombus.

If the two diagonals of a parallelogram are $72 \, cm$ and $30 \, cm$ respectively,then find its perimeter in $cm$.

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