In rhombus ABCD,AC = 16 and BD = 30. Find its | vedclass

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(A) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Then,$AO = OC =

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$68$

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(A) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Then,$AO = OC = \frac{AC}{2} = \frac{16}{2} = 8$ and $BO = OD = \frac{BD}{2} = \frac{30}{2} = 15$.In the right-angled triangle $	riangle AOB$,by the Pythagoras theorem:$AB^2 = AO^2 + BO^2$$AB^2 = 8^2 + 15^2$$AB^2 = 64 + 225 = 289$$AB = \sqrt{289} = 17$.The perimeter of a rhombus is $4 	imes 	ext{side}$.Perimeter $= 4 	imes 17 = 68$.

In rhombus $ABCD$,$AC = 16$ and $BD = 30$. Find its perimeter.

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