In rhombus ABCD,AB = 25 and AC = 14. Find BD. | vedclass

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(D) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Given $AB = 25$

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(D) In a rhombus,the diagonals bisect each other at right angles $(90^{\circ})$.Let the diagonals $AC$ and $BD$ intersect at point $O$.Given $AB = 25$ and $AC = 14$.Since diagonals bisect each other,$AO = OC = AC / 2 = 14 / 2 = 7$.In the right-angled triangle $	riangle AOB$,by the Pythagorean theorem:$AO^2 + OB^2 = AB^2$$7^2 + OB^2 = 25^2$$49 + OB^2 = 625$$OB^2 = 625 - 49 = 576$$OB = \sqrt{576} = 24$.Since $BD = 2 	imes OB$,we have $BD = 2 	imes 24 = 48$.

In rhombus $ABCD$,$AB = 25$ and $AC = 14$. Find $BD$.

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