Saturday, November 15, 2008

Elearn Geometry Problem 207: Right Triangle, Inradius, Exradius, Hypotenuse

Right triangle

See complete Problem 207 at:

Right Triangle, Hypotenuse, Inradius, Exradius relative to the hypotenuse. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1 comment:

  1. let s be the semi perimeter and r be the in radius and ra be the exradii of excircle relative to the side BC and let a,b,c be lenghths of sides BC,CA,AB we have some known relations AE=AD=s-a and AF=AG=s so EF=AF-AE=a and DG=AG-AD=a hence EF=DG=BC=a. we know that ra=Ar of TrABC/(s-a) and r=Ar of TrABC/s . so ra-r=Ar of TrABC.a/s(s-a)= Tan A/2 .a = Tan 45.a = a (Angle A = 90) HENCE ra-r=a1=a2=a. that means BC=EF=DG=ra-r