PQRS is a quadrilateral and the diagonals intersect at o. Show that PQ+QR+RS+SP is greater than PR+QS.
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The figure would be something like above.
To prove this, we will use the property of triangles that states that,
" Sum of any two sides of a triangle is greater than the third side"
In triangle PQS,
PQ+SP>QS------(1)
In triangle SRQ,
RS+QR>QS-------(2)
In triangle PQR,
PQ+QR>PR---------(3)
In triangle PSR,
SP+RS>PR-----------(4)
Adding (1),(2),(3) and (4), we get
PQ+SP+RS+QR+PQ+QR+SP+RS> QS+QS+PR+PR
2PQ+2QR+2RS+2SP>2QS+2PR
2(PQ+QR+RS+SP)>2(PR+QS)
Cancelling 2 from both sides,
PQ+QR+RS+SP>PR+QS
Hope it helps! Please mark it as 'Brainliest'!
To prove this, we will use the property of triangles that states that,
" Sum of any two sides of a triangle is greater than the third side"
In triangle PQS,
PQ+SP>QS------(1)
In triangle SRQ,
RS+QR>QS-------(2)
In triangle PQR,
PQ+QR>PR---------(3)
In triangle PSR,
SP+RS>PR-----------(4)
Adding (1),(2),(3) and (4), we get
PQ+SP+RS+QR+PQ+QR+SP+RS> QS+QS+PR+PR
2PQ+2QR+2RS+2SP>2QS+2PR
2(PQ+QR+RS+SP)>2(PR+QS)
Cancelling 2 from both sides,
PQ+QR+RS+SP>PR+QS
Hope it helps! Please mark it as 'Brainliest'!
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