Question

PQRS is a quadrilateral in which diagonals PR and QS intersect at O. Prove that PQ+QR+RS+SP>PR+QS. Please solve this question

Answer 1

Answer:

See,

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 

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