Question

# State True or FalseIf $$PQRS$$ is a quadrilateral in which diagonal $$PR$$ and $$QS$$ intersect at $$O$$. Then The sum of all the sides is greater than the sum of its diagonal i.e. $$PQ+QR+RS+SP>PR+QS$$

A
True
B
False

Solution

## The correct option is A True$$\Rightarrow$$  In triangle, sum of two sides is more than the third side.So, in $$\triangle PQR,$$$$\Rightarrow$$  $$PQ+QR>PR$$                ----- ( 1 )In $$\triangle PSR,$$$$\Rightarrow$$  $$PS+SR>PR$$                -----( 2 )Adding equation ( 1 ) and ( 2 ) we get,$$\Rightarrow$$  $$PQ+QR+RS+SP>2PR$$      ------ ( 3 )Now, in $$\triangle QRS,$$$$\Rightarrow$$  $$QR+RS>QS$$              ------ ( 4 )In $$\triangle QSP,$$$$\Rightarrow$$  $$PQ+SP>QS$$                 ----- ( 5 )Adding ( 4 ) and ( 5 ) we get,$$\Rightarrow$$  $$PQ+QR+RS+SP>2QS$$        ----- ( 6 )Adding ( 3 ) and ( 6 ) we get,$$\Rightarrow$$  $$2(PQ+QR+RS+SP)>2(PR+QS)$$$$\therefore$$  $$PQ+QR+RS+SP>PR+QS$$Mathematics

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