In fig. 12.116 , T is a point on the side QR of ∆ PQR and S is a point such that RT=St.
Prove that PQ+PR>QS
Asked by shivendrasinghfauzdar | 9th Oct, 2021, 11:30: AM
As sum of two sides of a triangle is greater than the third side, we have
PQ + PR > QR
Therefore, PQ + QR > QT + TR ... (As QR = QT + TR)
Therefore, PQ + QR > QT + TS ... (As TR = TS) (1)
Also, in triangle QST, QT + TS > QS ... (2)
From (1) and (2), we have
PQ + QR > QS
Answered by Renu Varma | 23rd Oct, 2021, 07:15: PM
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