In rhombus XYZW,XZ = 14 and YW = 48. Then,XY | vedclass

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(B) In rhombus $XYZW$,let the diagonals $XZ$ and $YW$ intersect at point $O$.The diagonals of a rhombus bisect each other at right angles.Therefore,$O

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(B) In rhombus $XYZW$,let the diagonals $XZ$ and $YW$ intersect at point $O$.The diagonals of a rhombus bisect each other at right angles.Therefore,$OX = \frac{1}{2} XZ = \frac{1}{2} 	imes 14 = 7$ and $OY = \frac{1}{2} YW = \frac{1}{2} 	imes 48 = 24$.In the right-angled triangle $\Delta XOY$,by the Pythagorean theorem:$XY^2 = OX^2 + OY^2$$XY^2 = 7^2 + 24^2$$XY^2 = 49 + 576 = 625$$XY = \sqrt{625} = 25$.

In rhombus $XYZW$,$XZ = 14$ and $YW = 48$. Then,$XY = \ldots \ldots \ldots$.

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