Question

PQRS is a quadrilateral in which diagnols PR and QS intersect at O.Prove that PQ+QR+RS+SP>PR+QR.​

Answer 1

Step-by-step explanation:

given that ,

PQRS is a quadrilateral in which diagonal PR and QS intersect at a O .

to prove - PQ +QR +RS+SP < 2 ( PR + QS )

proof -

we know that sum of any two side of a triangle is greater than the third side .

.'. in Δ PQO ,

PO+QO>PQ , .......................(i)

in Δ SOP

SO + PO >PS , .........................(ii)

in Δ SOR

SO + OR > RS ...........................(iii)

in Δ QOR ,

QO + OR > QR ...........................(iv)

on adding eqn. i , ii , iii & iv

we get ,

PO+QO+SO+PO+SO+OR+QO+OR > PQ+PS+SR+QR

also ⇒ 2 ( PO + QO + SO + OR ) > PQ+PS+SR + QR

= 2( PR + QS ) > PQ+PS+RS + QR ( proved) ....