Question

In Fig. PQR is a triangle and S is any point in its interior, show that SQ + SR < PQ + PR. [3 MARKS]

Answer 1

Process : 2 Marks
Proof : 1 Mark

Given: S is any point in the interior of $△$PQR

To prove: SQ + SR $<$ PQ + PR

Constructions: Produce QS to meet PR in T.

Proof:

In $△$ PQT, we have

PQ + PT $>$ QT
[Sum of two sides of a triangle are greater than the third side]

$\Rightarrow$ PQ + PT $>$ QS + ST ......(i) [$\because$ QT = QS + ST]

In $△$RST, we have

ST + TR $>$ SR .........(ii)

Adding (i) and (ii), we get

PQ + PT + ST + TR $>$ SQ + ST + SR

$\Rightarrow$ PQ + (PT + TR) $>$ SQ + SR

$\Rightarrow$ PQ +PR $>$ SQ + SR

$\Rightarrow$ SQ + SR $<$ PQ + PR.