tag:blogger.com,1999:blog-6933544261975483399.post7919285140430827006..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 203: Right Triangle, Hypotenuse,ExcirclesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-61321344890721983132012-08-13T16:00:25.221-07:002012-08-13T16:00:25.221-07:00Solution of problem 203.
Let s be the semiperimete...Solution of problem 203.<br />Let s be the semiperimeter of ABC.<br />From problem 202, rb = s - c, and rc = s – b, so<br />rb + rc = 2s – b - c =a + b + c – b –c = a.<br />Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68370894593649279392009-09-15T01:43:28.529-07:002009-09-15T01:43:28.529-07:00Let radius of circle(B) and circle(C) be b & c...Let radius of circle(B) and circle(C) be b & c respectively.<br /><br />Let 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.<br /><br />Now tangents from point C to circle(C) will be equal<br />x+a = y+b+c------equation1<br />Also tangents from point B to circle(B) will be equal<br />b+c+x = a+y ------equation2<br /><br />Adding 1 & 2 you get x=y. You put it in any of the above equation you will get a = b+c.shaileshhttps://www.blogger.com/profile/13003995866362418349noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21066144168463836732009-09-03T03:17:01.486-07:002009-09-03T03:17:01.486-07:00let S be the area of right triangle ABC and p be t...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=avijay9290009015http://vijay.comnoreply@blogger.com