Tuesday, October 21, 2008

Elearn Geometry Problem 193

Area of a Triangle, Semiperimeter, Inradius

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.

3 comments:

  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

    ReplyDelete
  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.

    ReplyDelete
  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

    ReplyDelete