Question

Let S be any point in the interior of ∆PQR. Prove that the perimeter of ∆SQR is less than the perimeter of∆PQR.

Answer 1

As S in interior of ∆PQR
so ∆SQR will be smaller triangle than ∆PQR.
And perimeter is the sum of all side , now as ∆SQR is smaller so it's length SQ and length SR will be smaller than PQ and PR .
thus perimeter of ∆SQP will be less than of ∆PQR

perimeter of ∆SQR=SQ+SR+QR
perimeter of ∆PQR=PQ+PR+QR
here QR is same in both triangle.

Answer 2

Answer in attached photo