Question 6
S is any point on side QR of a ΔPQR. Show that PQ + QR + RP > 2 PS.
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Solution
Given in ΔPQR, S is any point on side QR
InΔPQS,PQ+QS>PS
[sum of two sides of a trianlge is greater than the third side] Similarly,inΔPRS,SR+PR>PS
[sum of two sides of a triangle is greater than the third side]
on adding Eqs (i) and (ii). we get, PQ+QS+SR+RP>2PS ⇒PQ+(QS+SR)+RP>2PS⇒PQ+QR+RP>2PS[∵QR=QS+SR]