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S is any point interior of triangle PQR, prove that PQ+PR>QS+Rs
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I think you have an error in ur question... it should be SQ instead of PQ.. and if it is so ... .
here's ur answer...
In triangle STR,
ST + TR > SR …….(i) (Sum of two sides of a triangle is greater than the third side)
In triangle PQT,
PQ + PT > QT …..(ii) (Sum of two sides of a triangle is greater than the third side)
Adding (i) and (ii), we get
ST + TR + PQ + PT > SR + QT
PQ + PR + ST > SR + QS + ST (PT + TR = PR and QT = QS + ST)
PQ + PR > SQ + SR
Hence proved.
figure in attachment...
hope it helps. :))
here's ur answer...
In triangle STR,
ST + TR > SR …….(i) (Sum of two sides of a triangle is greater than the third side)
In triangle PQT,
PQ + PT > QT …..(ii) (Sum of two sides of a triangle is greater than the third side)
Adding (i) and (ii), we get
ST + TR + PQ + PT > SR + QT
PQ + PR + ST > SR + QS + ST (PT + TR = PR and QT = QS + ST)
PQ + PR > SQ + SR
Hence proved.
figure in attachment...
hope it helps. :))
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umang100010:
Thank you so much....
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