In rhombus ABCD,AB = 13 and AC = 24. Find BD. | vedclass

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(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Since $AC = 24$

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Since $AC = 24$,the length of $AO = AC / 2 = 24 / 2 = 12$.In the right-angled triangle $	riangle AOB$,by the Pythagorean theorem,$AB^2 = AO^2 + BO^2$.Given $AB = 13$ and $AO = 12$,we have $13^2 = 12^2 + BO^2$.$169 = 144 + BO^2$.$BO^2 = 169 - 144 = 25$.$BO = \sqrt{25} = 5$.Since the diagonals bisect each other,$BD = 2 	imes BO = 2 	imes 5 = 10$.

In rhombus $ABCD$,$AB = 13$ and $AC = 24$. Find $BD$.

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