See complete Problem 206 at:
gogeometry.com/problem/p206_right_triangle_area_inradius.htm
Area of a Right Triangle Area, Inradius, Exradius relative to the hypotenuse. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Saturday, November 15, 2008
Elearn Geometry Problem 206: Area of a Right Triangle, Inradius, Exradius
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Solution of problem 206.
ReplyDeleteLet s be the semi perimeter of triangle ABC. The area of this triangle is equal to S = s.r. By problem 201, we have ra = s. So S = r.ra.
Let AB=a, AC=b and area right triangle ABC=s
ReplyDeleteBy Poncelet we have: a+b=c+2r
(ra-r)+(ra-r)=a+b-2r
Simplifying
ra+r=a+b
Now we know that
(ab)/2=(a-r)(b-r)
s=(a-r)(b-r)
Rewriting this expression
r^2=r(a+b)-s................(1)
ra+r=a+b can be rewrite this way: rar+r^2=r(a+b)
rar+r^2=r(a+b)...............(2)
Now
rar+r(a+b)-s=r(a+b)
s=rar
Q.E.D.
By Tony Garcia