Question

In the given figure, PQR is a triangle and S is any point in its interior. Then which of the following option 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]
$\Rightarrow PQ+PT>SQ+ST$ $\dots$ (1)
$[\because QT=SQ+ST]$
Also, in $\Delta RST$,
$ST+TR>SR$ $\dots$ (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.]
$\therefore PQ+PR>SQ+SR$