Question

a circle with Centre P is inscribed in the triangle ABC side a b side BC and side AC touch the circle at points l m and n respectively radius of the circle is a prove that area of triangle ABC is equal to half of( AB + BC + AC )x r

Answer 1

Answer:

Step-by-step explanation:

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

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

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