tag:blogger.com,1999:blog-6933544261975483399.post4499976401771979011..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 206: Area of a Right Triangle, Inradius, ExradiusAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-1983593797115448892013-01-29T08:56:25.135-08:002013-01-29T08:56:25.135-08:00Let AB=a, AC=b and area right triangle ABC=s
By Po...Let AB=a, AC=b and area right triangle ABC=s<br />By Poncelet we have: a+b=c+2r<br />(ra-r)+(ra-r)=a+b-2r<br />Simplifying<br />ra+r=a+b<br />Now we know that <br />(ab)/2=(a-r)(b-r)<br />s=(a-r)(b-r)<br />Rewriting this expression<br />r^2=r(a+b)-s................(1)<br />ra+r=a+b can be rewrite this way: rar+r^2=r(a+b)<br />rar+r^2=r(a+b)...............(2)<br />Now <br />rar+r(a+b)-s=r(a+b)<br />s=rar<br />Q.E.D.<br />By Tony GarciaAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-46053775048460814842012-08-18T06:49:12.442-07:002012-08-18T06:49:12.442-07:00Solution of problem 206.
Let s be the semi perimet...Solution of problem 206.<br />Let 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.<br />Nilton Lapanoreply@blogger.com