In rhombus ABCD,AC = 12 and BD = 16. Find the | vedclass

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(C) The diagonals of a rhombus bisect each other at right angles.Let the diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.Then,$m\

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$40$

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(C) The diagonals of a rhombus bisect each other at right angles.Let the diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $M$.Then,$m\angle AMB = 90^{\circ}$.Since the diagonals bisect each other,$AM = \frac{1}{2} AC = \frac{1}{2}(12) = 6$ and $BM = \frac{1}{2} BD = \frac{1}{2}(16) = 8$.In right-angled $\Delta AMB$,by the Pythagorean theorem:$AB^2 = AM^2 + BM^2$$AB^2 = 6^2 + 8^2$$AB^2 = 36 + 64 = 100$$AB = \sqrt{100} = 10$.The perimeter of a rhombus is $4 	imes 	ext{side length}$.Perimeter $= 4 	imes AB = 4 	imes 10 = 40$.Thus,the perimeter of rhombus $ABCD$ is $40$.

In rhombus $ABCD$,$AC = 12$ and $BD = 16$. Find the perimeter of rhombus $ABCD$.

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