The perimeter of rhombus ABCD is 68. If AC = | vedclass

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(A) rhombus is a quadrilateral where all sides are equal. Let the side length be $s$.Given the perimeter is $68$,we have $4s = 68$,which implies $s =

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$16$

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(A) rhombus is a quadrilateral where all sides are equal. Let the side length be $s$.Given the perimeter is $68$,we have $4s = 68$,which implies $s = 17$.In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Since $AC = 30$,the length $AO = AC / 2 = 15$.In the right-angled triangle $	riangle AOB$,by the Pythagorean theorem,$AO^2 + BO^2 = AB^2$.Substituting the values: $15^2 + BO^2 = 17^2$.$225 + BO^2 = 289$.$BO^2 = 289 - 225 = 64$.$BO = \sqrt{64} = 8$.Since the diagonals bisect each other,$BD = 2 	imes BO = 2 	imes 8 = 16$.

The perimeter of rhombus $ABCD$ is $68$. If $AC = 30$,find $BD$.

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