S is any point on the side QR of a ∆ PQR. Show that: PQ + QR + RP > 2 PS.
Solution:
Given, PQR is a triangle
S is any point on the side QR of the triangle
We have to show that PQ + QR + RP > 2 PS
We know that in a triangle the sum of two sides of a triangle is greater than the third side.
Considering triangle PQS,
PQ + QS > PS ------------ (1)
Considering triangle PRS,
SR + RP > PS ------------- (2)
On adding (1) and (2),
PQ + QS + SR + RP > PS + PS
PQ + (QS + SR) + RP > 2 PS
From the figure,
QR = QS + SR
Therefore, PQ + QR + RP > 2 PS
✦ Try This: In ΔABC, DE∣∣BC so that AD=(7x-4)cm, AE=(5x-2)cm, DB=(3x+4)cm and EC=3x. Then, we have
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.3 Problem 6
S is any point on the side QR of a ∆ PQR. Show that: PQ + QR + RP > 2 PS
Summary:
S is any point on the side QR of a ∆ PQR. It is shown that PQ + QR + RP > 2 PS
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