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Question

In the following figure, the inscribed circle of ΔABC with centre P touches the sides AB, BC and AC at points L, M, N respectively. Show that A(ΔABC)=12× (perimeter of ΔABC) × (radius of inscribed circle).
599735_03d6a3fa56424ee78769a1bd615db5a1.png

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Solution

Given: The inscribed circle of ΔABC touches side AB at L, side BC at M and side AC at N.

Proof: By Tangent theorem
Radius PL AB, PM BC and PN AC
In ΔAPBA(ΔAPN)=12×b×h
=12×AB×r ....... (1)
In ΔBPCA(ΔBPC)=12×BC×r ....... (2)
In ΔAPCA(ΔAPC)=12×AC×r ........ (3)
Adding (1), (2) and (3)
A(ΔAPB)+A(ΔBPC)+A(ΔAPC)=12×AB×r+12×BC×r+12×AC×r
=12(AB+BC+AC)×r
A(ΔABC)=12(perimeter ofΔABC)×radius of inscribed circle

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