Question

In figure 3.60, point S is any point on side QR of ΔPQR Prove that : PQ+QR+RP>2PS

Answer 1

Solution:-

given by:- point S is any point on side QR of ΔPQR.

prove:- a triangle sum of two side is always greater than the third side,
so in triangle PQS we have,
PQ+QS+PS .................(1)
similarly ,
in triangle PSR we have,
PR+SR+PS ...............(2)

ADDINTING BOTH SIDE (1) AND (2)
we get,
( PQ+ QS+PR+SP)>2PS
=> (PQ+PR+QS+SP)>2PS
=>[ PQ+PR+QR > 2PA ] PROVED


■I HOPE ITS HELP■

Answer 2

Answer:

PQ+QR+RP>2PS

Step-by-step explanation:

given by:- point S is any point on side QR of ΔPQR.

prove:- a triangle sum of two side is always greater than the third side,

so in triangle PQS we have,

PQ+QS+PS .................(1)

similarly ,

in triangle PSR we have,

PR+SR+PS ...............(2)

ADDINTING BOTH SIDE (1) AND (2)

we get,

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

=> (PQ+PR+QS+SP)>2PS

=>[ PQ+PR+QR > 2PA

Hence proved