Question

A circle with centre p is inscribed in the ∆ABC. Side AB,side BC and side AC touch the circle at points L,M and N respectively. Radius of the circle is r
Prove that :

A(∆ABC)=1:2(AB+BC+AC)×r

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Consider $P$ is the center of the circle.  

Radius of the circle =$r$

Prove that

$\text{Area of }\Delta ABC=\frac{1}{2}(AB+BC+AC)\times r$

Draw lines between $PA, \text{ }PB$ and $PC$

AB is tangent on the circle, touch the circle at the point L, and PL is the radius.

According to the tangent radius theory $PL\perp AB$

therefore we can proved that  

$PN\perp AC$

$PM\perp BC$

We know that,

$PL=PM=PN=r\text{ (Radius of the circle)}$

And $\Delta APL, \Delta APN, \Delta BPL, \Delta BPM, \Delta CPM \text{ and }\Delta CPN\text{ are right angled triangle}$

Therefore,

$\text{Area of }\Delta ABC=\text{Area of }\Delta APL+\text{Area of }\Delta APN+\text{Area of }\Delta BPL+\text{Area of }\Delta BPM+\text{Area of }\Delta CPM+\text{Area of }\Delta CPN\\\\=\frac{1}{2}\times PL\times AL+\frac{1}{2}\times PN\times AN+\frac{1}{2}\times PL\times BL+\frac{1}{2}\times PM\times BM+\frac{1}{2}\times PM\times CM+\frac{1}{2}\times PN\times CN$

$=\frac{1}{2}\times r\times AL+\frac{1}{2}\times r\times AN+\frac{1}{2}\times r\times BL+\frac{1}{2}\times r\times BM+\frac{1}{2}\times r\times CM+\frac{1}{2}\times r\times CN\\\\=\frac{1}{2}\times r(AL+AN+BL+BM+CM+CN)\\\\=\frac{1}{2}\times r(AL+BL+BM+CM+CN+AN)\\\\=\frac{1}{2}\times r(AB+BC+AC)$