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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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,$XY = \ldots \ldots \ldots$.

Similar Questions

In $\Delta PQR$,$\overline{PS}$ is a median. If $PQ = 9$,$PR = 40$ and $PS = 20.5$,find $QR$.

In the figure,if $DE \parallel BC,$ find the ratio of $\operatorname{ar}(ADE)$ and $\operatorname{ar}(DECB).$

In $\Delta ABC$ and $\Delta PQR$,$\angle A \cong \angle P$ and $\angle B \cong \angle R$. Then,the correspondence $ABC \leftrightarrow \ldots$ is a similarity.

In $\Delta PQR$,$m \angle P = 90^{\circ}$ and $\overline{PM}$ is an altitude. If $QM = 2x^{2}$ and $RM = 8x^{2}$,then $PM = \ldots \ldots$

In rectangle $ABCD$,$AC = 25$ and $CD = 7$. Then,the perimeter of the rectangle is........

Vedclass Test Series

Exam Paper Generator

Online Exam Module