Question

in fig 3.60 point s is any point on side QR of Δ PQR prove that PQ+QR+RP>2PS

Answer 1

Answer:

Theorem: In a triangle , sum of the length of any two sides is greater than the third side.

Draw the triangle and join the points P and S

In the figure,

In  ΔPQS, according to the theorem,

PQ+QS>PS  ........... (1)

In  ΔPSR, according to the theorem,

PR+SR>PS  ........... (2)

Adding (1) and (2),

PQ+QS+SR+PR>PS+PS

PQ+(QS+SR)+PR>2PS

PQ+QR+PR>2PS

solution

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