Question

In the given figure, PQR is a triangle and S is any point in its interior. Which of the following options is correct?

• A. SQ + SR < PQ + PR
• B. SQ + SR > PQ + PR
• C. SQ + PQ > SR + PR
• D. SQ - SR > PQ - PR

Answer 1

The correct option is A SQ + SR < PQ + PR

S is any point in the interior of $\Delta PQR$.
$In\Delta PQT,PQ+PT>QT[\because$ Sum of the two sides of a triangle is greater than the third side]
$\Rightarrow PQ+PT>SQ+ST$
$[\because QT=SQ+ST]...\left(1\right)$
Also, in $\Delta RST$,
ST + TR > SR...(2)$[\because$ Sum of the two sides of a triangle is greater than the third side]
Adding (1) and (2), we get
PQ + PT + ST + TR > SQ + ST + SR
$\Rightarrow PQ+(PT+TR)>SQ+SR$
[Cancelling ST on both sides.]
$\therefore PQ+PR>SQ+SR$