Question

if PQRS is a square in which diagonals PR and QS intersect at point O .then prove that -- PQ+QR+RS+SP< 2(PR + QS)

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

Solution :- In triangle POQ,

OP + OQ > PQ __(1)

Since sum of two sides is greater than 3rd sides.

In triangle QOR,

OQ + OR > QR ___(2)

Since sum of two sides is greater than 3rd sides.

In triangle SOR,

OS + OR > RS ___(3)

Since sum of two sides is greater than 3rd sides.

In triangle POS,

OP + OS > PS ___(4)

Since sum of two sides is greater than 3rd sides.

On adding all 4 equations we get,

OP + OQ + OQ + OR + OR + OS + OS + OP > PQ + PR + RS + PS

( OP + OR ) + ( OP + OR ) + ( OQ + OS ) + ( OQ + OS ) > PQ + PR + RS + PS

PR + PR + QS + QS > PQ + PR + RS + PS

PQ + PR + RS + PS > 2 (PR+QS)

$\textbf{ \large{ Q.E.D }}$

$\textbf{ Have great future ahead! }$