Question

23. In the given figure, S is any point on QR. Prove that
PQ + QR + RP > 2PS

Answer 1

Step-by-step explanation:

In ΔPQS

Sum of any two sides is greater than the third side.

So, PQ + QS > PS ......(1)

In ΔPSR

Sum of any two sides is greater than the third side.

So, PR + SR > PS ...(2)

Adding equations (1) and (2)

PQ + PR + QS + SR > PS + PS

PQ + PR + QR > 2PS [ QS + SR = QR]

Answer 2

Answer:

Step-by-step explanation:

In triangle PQS,

PQ+QS>PS   (1) (sum of two sides of a triangle is always greater than third side)

In triangle PSR,

SR+RP>PS    (2) (sum of two sides of a triangle is always greater than third side)

from (1) and (2)

PQ+QS+SR+RP>PS+PS

thus, PQ+QR+RS>2PS

Hence Proved