See complete Problem 193 at:

www.gogeometry.com/problem/p193_area_of_a_triangle_semiperimeter_inradius.htm

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

Post your solutions or ideas in the comments.

## Tuesday, October 21, 2008

### Elearn Geometry Problem 193

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This is easily done by remembering that:

ReplyDeleteTr. 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

if S denotes the area of the triangle ABC, then

ReplyDeleteS = 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.

Drop the radii from incentre at the points of contact.. We have three kites.

ReplyDeleteArea of ABC = r(s − a) + r(s − b) + r(s − c) = r(s − a + s − b + s − c) = r(3s − 2s) = rs

A=A1+A2+A3

ReplyDeleteA=(1/2)ar+(1/2)br+(1/2)cr

A=(1/2)r(a+b+c)

A=r.s