Question

UR
PQRS is a Rhombus. Prove that PR2 + QS2 = 4PQ2.​

Answer 1

Step-by-step explanation:

Let PQRS be a rhombus whose diagonals PR and QS intersect at O.

It is the property of a rhombus that diagonals perpendicularly bisect each other.

∴ In ΔPOQ,

PQ² = OP² + OQ²

$(\frac{PR}{2})^{2}$ + $(\frac{QS}{2})^{2}$ = PQ²

$\frac{PR^{2} }{4}$ + $\frac{QS^{2} }{4}$ = PQ²

$\frac{PR^{2} + QS^{2} }{4}$ = PQ²

PR² + QS² = 4PQ²

Hence Proven

Hope this helps :)