Question

S is any point in the interior of Δ PQR. Show that SQ + SR < PQ + PR

Answer 1

produce QS to intresect PR at T
from Δ PQT, we have
PQ+PT>QT


PQ+PT>SQ+ST----(1)
ST+TR>SR----------(2)

ADDING (1) AND (2) WE GET,
PQ+PT+ST+TR>SQ+ST+SR
PQ+PT+TR>SQ+SR
PQ+PR>SQ+SR


SQ+SR<PQ+PR

Answer 2

Thank you for asking this question:

Here is your answer:

First of all we will draw QS so that it can intersect PR at T

From ΔPQT we will have this:

PQ + PT > QT  (this is the sum of any two sides which are comparatively greater than third side)

PQ + PT > SQ + ST

From Δ TSR we will have this:

ST + TR > SR

Now we will add both the equations

PQ + PT  + ST + TR > SQ + ST + SR

PQ + PT + TR > SQ + SR

PQ + PR > SQ + SR

If there is any confusion please leave a comment below.