Question

prove that area of triangle = 1/2 (perimeter of triangle) *r​

Answer 1

Sol:

Given: A circle touching the side BC of ΔABC at P and AB, AC produced at Q and R respectively.

RTP: AP = 1/2 (Perimeter of ΔABC)

Proof: Lengths of tangents drawn from an external point to a circle are equal.

⇒ AQ = AR, BQ = BP, CP = CR.

Perimeter of ΔABC = AB + BC + CA

= AB + (BP + PC) + (AR – CR)

= (AB + BQ) + (PC) + (AQ – PC) [AQ = AR, BQ = BP, CP = CR]

= AQ + AQ

= 2AQ

⇒ AQ = 1/2 (Perimeter of ΔABC)

∴ AQ is the half of the perimeter of ΔABC.

Hope it helps

Please mark it brainliest