Tuesday, October 21, 2008

Elearn Geometry Problem 194

Area of a Triangle, Semiperimeter, Exradius

See complete Problem 194 at:
www.gogeometry.com/problem/p194_area_of_a_triangle_semiperimeter_exradius.htm

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

Post your solutions or ideas in the comments.

5 comments:

  1. Hints:
    Area ABC = Area ABE + Area ACE - Area BCE.
    Apply Area of a triangle = bh/2 three times.

    ReplyDelete
  2. Let the perpendicular from E_a meet AC extended in M and from E_a to BC in N. We know that AM = s. Therefore Tr. ABC = s*r_a - (CN*r_a+BN*r_a)
    = s*r_a - (CN+BN)*r_a
    = s*r_a - BC * r_a
    =(s-a)r_a
    Ajit: ajitathle@gmail.com

    ReplyDelete
  3. I was wondering if somone could give more detail on either of these proofs. thanksssss

    ReplyDelete
  4. If Q is the point of tangency for the extension of the line AC with the circle r_a , then the length of the AQ =s by problem 140 .

    Let O be the center of the circle with radius r inscribed onto the triangle ABC. And let point M be the point of tangency for circle r with the line AC.
    And let point N be the point of tangency for circle r with the line AB.

    Then we can derive an equation for AM :

    AM =AN

    therefore MC = b-AM and NB =c-AM

    BC =MC+NB
    a=b-AM+c-AM
    a=b+c-2AM
    2AM=-a+b+c
    2AM+2a=a+b+c
    2(AM+a)=2s

    (1.) AM=s-a

    By similar triangles E_aQA and OMA we can get an equation for r

    OM/AM=E_aQ/AQ
    r/(s-a)=r_a/s

    (2.) r_a=rs/(s-a)


    In problem 193 we see that the area of triangle ABC S=rs. Substituting equation 2 in this equation we get :

    S=rs
    S=r_as(s-a)/s
    S=r_a(s-a)

    ReplyDelete
    Replies
    1. It appears i made a mistake at equation 2 .
      It should be rewritten as:

      r=r_a(s-a)/s

      And then substituted into S=rs

      Delete