The perimeter of rhombus ABCD is 116. If AC = | vedclass

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(D) $1$. The perimeter of a rhombus is given by $4 	imes 	ext{side}$. Let the side length be $s$. So, $4s = 116$, which gives $s = 29$.$2$. In a rho

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$40$

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(D) $1$. The perimeter of a rhombus is given by $4 	imes 	ext{side}$. Let the side length be $s$. So, $4s = 116$, which gives $s = 29$.$2$. In a rhombus, the diagonals bisect each other at right angles $(90^{\circ})$.$3$. Let the diagonals $AC$ and $BD$ intersect at point $O$. Then $AO = AC / 2 = 42 / 2 = 21$.$4$. In the right-angled triangle $	riangle AOB$, by the Pythagorean theorem: $AO^2 + BO^2 = AB^2$.$5$. Substituting the values: $21^2 + BO^2 = 29^2$.$6$. $441 + BO^2 = 841$.$7$. $BO^2 = 841 - 441 = 400$.$8$. $BO = \sqrt{400} = 20$.$9$. Since the diagonals bisect each other, $BD = 2 	imes BO = 2 	imes 20 = 40$.

The perimeter of rhombus $ABCD$ is $116$. If $AC = 42$, find $BD$.

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