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.

DeleteDo you have similar blog/site for other maths topics? Your blog is excellent.

ReplyDelete