Question

S is any point in the interior of triangle PQR.Show that the (PQ+PR) >(SQ+SR)

Answer 1

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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.

Answer 2

Draw QS so that it can intersect PR at T

In ΔPQT,

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

PQ + PT > SQ + ST - 1

In Δ TSR,

ST + TR > SR - 2

Now, 1 + 2

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

PQ + PT + TR > SQ + SR

PQ + PR > SQ + SR