Question

PQRS is a quadrilateral is PQ + QR + RS + SP > than 2 (PR + SQ).​

Answer 1

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) ….

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