Tuesday, October 21, 2008

Elearn Geometry Problem 194

See complete Problem 194 at:

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

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

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

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

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)

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