S is any point in the interior of ∆ PQR. Show that SQ + SR < PQ + PR.
Solution:
Given, PQR is a triangle.
S is any point in the interior of triangle PQR
We have to show that SQ + SR < PQ + PR
Extend QS to meet PR at T.
We know that in a triangle the sum of any two sides is greater than the third side
Considering triangle PQT,
PQ + PT > QT
From the figure,
QT = SQ + ST
Now, PQ + PT > SQ + ST -------------- (1)
Considering triangle TSR,
ST + TR > SR ------------------------------- (2)
Adding (1) and (2),
PQ + PT + ST + TR > SQ + ST + SR
Canceling common term,
PQ + PT + TR > SQ + SR
From the figure,
PR = PT + TR
So, PQ + PR > SQ + SR
Therefore, SQ + SR < PQ + PR
✦ Try This: In figure, Q is a point on side SR of ∆PSR such that PQ = PR. Prove that PS > PQ
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Sample Problem 4
S is any point in the interior of ∆ PQR. Show that SQ + SR < PQ + PR
Summary:
S is any point in the interior of ∆ PQR. It is shown that SQ + SR < PQ + PR
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