diagonals PR,SQ of an quadrilateral PQRS meet at o provethat PQ+QR+RS+SP>(PR+QS)
Asked by | 8th Sep, 2012, 12:46: PM
In a triangle, sum of two sides is more the third side.
So, in triangle PQR;
PQ + QR > PR
(1)
In triangle PSR;
PS + SR > PR
(2)
Adding (1) and (2);
PQ + QR + RS + SP > 2PR
(3)
Similarly, it can be proved that
PQ + QR + RS + SP > 2QS
(4)
Adding (3) and (4);
2(PQ + QR + RS + SP) > 2(PR + QS)
i.e. PQ + QR + RS + SP > PR + QS
Hence, proved.

In a triangle, sum of two sides is more the third side.
So, in triangle PQR;
PQ + QR > PR (1)
In triangle PSR;
PS + SR > PR (2)
Adding (1) and (2);
PQ + QR + RS + SP > 2PR (3)
Similarly, it can be proved that
PQ + QR + RS + SP > 2QS (4)
Adding (3) and (4);
2(PQ + QR + RS + SP) > 2(PR + QS)
i.e. PQ + QR + RS + SP > PR + QS
Hence, proved.

Answered by | 8th Sep, 2012, 08:27: PM
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