PQRS is a square. If PR = 10 , cm,then the ar | vedclass

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

Question

(B) Let the side of the square $PQRS$ be $a \, cm$.In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a\sqrt{2}$.Given that t

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) Let the side of the square $PQRS$ be $a \, cm$.In a square,the diagonal $d$ is related to the side $a$ by the formula $d = a\sqrt{2}$.Given that the diagonal $PR = 10 \, cm$,we have $a\sqrt{2} = 10$.Solving for $a$,we get $a = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, cm$.The area of a square is given by the formula $	ext{Area} = a^2$.Substituting the value of $a$,we get $	ext{Area} = (5\sqrt{2})^2 = 25 	imes 2 = 50 \, cm^2$.Alternatively,the area of a square can be calculated using the diagonal $d$ as $	ext{Area} = \frac{1}{2} 	imes d^2$.Substituting $d = 10 \, cm$,we get $	ext{Area} = \frac{1}{2} 	imes (10)^2 = \frac{1}{2} 	imes 100 = 50 \, cm^2$.

$PQRS$ is a square. If $PR = 10 \, cm$,then the area of $PQRS = \dots \, cm^2$.

Similar Questions

In quadrilateral $ABCD$,$\angle A = 100^{\circ}$,$\angle B = 80^{\circ}$,and $\angle C = 120^{\circ}$. Find $\angle D$. (in $^{\circ}$)

In quadrilateral $ABCD$, one of its diagonals $AC$ measures $20 \, cm$. The altitudes on $AC$ from vertices $B$ and $D$ are $8 \, cm$ and $12 \, cm$ respectively. Find the area of quadrilateral $ABCD$ in $cm^2$.

Diagonals of a rectangle are equal and perpendicular. Is this statement true? Give reason for your answer.

In parallelogram $ABCD$,diagonals $AC$ and $BD$ intersect at point $O$. Point $P$ lies on line segment $BO$. Prove that,$ar(ADO) = ar(CDO)$.

If $APB$ and $CQD$ are two parallel lines,then the bisectors of the angles $APQ, BPQ, CQP$ and $PQD$ form

Vedclass Test Series

Exam Paper Generator

Online Exam Module