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Question

State True or False
If $$PQRS$$ 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
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B
False
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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,
$$\Rightarrow$$  $$2(PQ+QR+RS+SP)>2(PR+QS)$$
$$\therefore$$  $$PQ+QR+RS+SP>PR+QS$$

1259028_1065301_ans_bb53db1d0d9a4be1944038976f812027.jpeg

Mathematics

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