See complete Problem 195 at:

www.gogeometry.com/problem/p195_area_of_a_triangle_inradius_exradius.htm

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

Post your solutions or ideas in the comments.

## Tuesday, October 21, 2008

### Elearn Geometry Problem 195

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lengths BC=a, CA=b, AB=c; s=(a+b+c)/2

ReplyDeleteS=sr

S=[EAB]+[EAC]-[EBC]=c(r_a)/2+b(r_a)/2-a(r_a)/2=(s-a)(r_a),=(s-b)(r_b)=(s-c)(r_c)

By Heron's formula:

S=√[s(s-a)(s-b)(s-c)]

S√[r(r_a)(r_b)(r_c)]=√[sr(s-a)(r_a)(s-b)(r_b)(s-c)(r_c)]=S²

S=√[r(r_a)(r_b)(r_c)]

let BC = a, AB = c and AC = b and s=(a+b+c)/2

ReplyDeleteuse the formulae, ra = S/(s-a), rb = S/(s-b) and rc = S/(s-c) and finally r = S/s where S denotes the area of the triangle ABC.

now r*ra*rb*rc = (S*S*S*S)/s(s-a)(s-b)(s-c)

but s(s-a)(s-b)(s-c) = S^2 and this follows from heron's formula.

which implies that r*ra*rb*rc = S^4/S^2 = S^2.

so that S = (r*ra*rb*rc)^(1/2).

Q. E. D.

To Antonio: When I try to get the comments or solutions of problem 196, I only get those of problem 195. Is there any bug in the links?

ReplyDeleteTo Nilton (problem 196), thanks for your information, the link has been corrected.

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