See complete Problem 203 at:
gogeometry.com/problem/p203_right_triangle_excircles_hypotenuse.htm
Right Triangle, Excircles, Exradii, Hypotenuse. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, November 10, 2008
Elearn Geometry Problem 203: Right Triangle, Hypotenuse,Excircles
See complete Problem 203 at:
gogeometry.com/problem/p203_right_triangle_excircles_hypotenuse.htm
Right Triangle, Excircles, Exradii, Hypotenuse. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
let S be the area of right triangle ABC and p be the semi perimeter then we have Tan A/2=S/p(p-a) but here Angle A=90 so S/p(p-a)= Tan 45 i so S=P(p-a) and so p(p-a)=(p-b)(p-c), and we have rb=S/(p-b)and rc=S/(p-c) so rb+rc=S(2p-b-c)/(p-b)(p-c)=S(a)/(p-b)(p-c)=a HENCE rb+rc=a
ReplyDeleteLet radius of circle(B) and circle(C) be b & c respectively.
ReplyDeleteLet tangent from point B to circle(C) be of length x and also let tangent from point C to circle(B) be of lenght y.
Now tangents from point C to circle(C) will be equal
x+a = y+b+c------equation1
Also tangents from point B to circle(B) will be equal
b+c+x = a+y ------equation2
Adding 1 & 2 you get x=y. You put it in any of the above equation you will get a = b+c.
Solution of problem 203.
ReplyDeleteLet s be the semiperimeter of ABC.
From problem 202, rb = s - c, and rc = s – b, so
rb + rc = 2s – b - c =a + b + c – b –c = a.