## Tuesday, October 21, 2008

### Elearn Geometry Problem 193

See complete Problem 193 at:

Area of a Triangle, Semiperimeter, Inradius. Level: High School, SAT Prep, College geometry

1. This is easily done by remembering that:
Tr. ABC = Tr. AIC + Tr. CIB + Tr. BIA
= r*b/2 + r*a/2 + r*c/2
= r(a+b+c)/2 = r.s
Ajit: ajitathle@gmail.com

2. if S denotes the area of the triangle ABC, then
S = area of triangles AIC+BIC+AIB
but area of triangleAIC=(1/2)r*b where b denotes the length of side AC. similarly, area of triangleBIC=(1/2)r*a and area of triangleAIB=(1/2)r*c where a and c denote the lengths of sides BC and AB respectively. it is easy to note that the inradius is perpendicular to the sides as they are tangents to the incircle.
thus adding the above results we get,
S = (1/2)r*(a+b+c) = r*(a+b+c)/2 = r*s where s denotes the semiperimeter of triangle ABC which equals (a+b+c)/2.
Q. E. D.

3. Drop the radii from incentre at the points of contact.. We have three kites.
Area of ABC = r(s − a) + r(s − b) + r(s − c) = r(s − a + s − b + s − c) = r(3s − 2s) = rs

4. A=A1+A2+A3
A=(1/2)ar+(1/2)br+(1/2)cr
A=(1/2)r(a+b+c)
A=r.s