Question

14. In quadrilateral PQRS, the diagonals PR and QS intersect each other at O. Show that
a. PQ+ QR + RS + SP > PR + QS
b. PQ + QR + RS + SP < 2(PR + QS)​

Answer 1

Answer:

(a)Two sides of a triangle are together greater than the third side.

Step-by-step explanation:

By this theorem,

PQ+QR>PR

and

PQ+PS>QS

Adoing these two expressions of inequality

we get

PQ+QR+RS+SP>PR+QS

ANSWER completed.

(a) This one is tough for me. Sorry.