Question

The diagonals PR and QS of quadrilateral PQRS intersect each other at point O. Prove that PQ+QR+RS+SP>PR+QS

Answer 1

In the quadrilateral PQRS ,join P to R andQ to S then,
In triangle PQR,
PR<PQ+QR ........(1)
(because in a triangle the sum
of two sides is greater than
the third side)
similarly,
QS<RS+SP .........(2)
From 1st and 2nd. we have,
PR+QS<PQ+QR+RS+SP
Hence proved

Answer 2

$\huge \underline\mathcal {soution : - ✌}$

Construct PM⊥QS and RN⊥QS

ar(ΔPSA)x ar(ΔQAR)

$= ( \frac{1}{2} \times AS \times PM)( \frac{1}{2} \times AQ \times RN)$

$= ( \frac{1}{2} \times RN \times AS )( \frac{1}{2} \times PM \times AQ)$

=ar(ΔSAR) x ar(ΔPAQ)

#hope it helps.

(refer to the attachment for the figure)