Answers

2014-12-25T17:19:45+05:30


Let the circle touches the sides BC, CA, AB of the right triangle ABC(right angled at C) at D, E and F respectively, where BC= a, CA=b , AB= c respectively

Since lengths of tangents drawn from an external point are equal

Therefore, AE=AF, and BD=BF

Also CE=CD=r

and b-r=AF , a- r= BF

Therefore AB=AF+BF

c= b-r + a-r AB=c=AF+BF=b-r+a-r

hence, r=a+b-c/2


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2014-12-25T18:45:38+05:30

Let the circle touches the sides  AB,BC and CA of triangle ABC at D, E and F 

Since lengths of tangents drawn from an external point are equal  We have

 AD=AF,  BD=BE  and CE=CF

Similarly EB=BD=r

Then we have  c = AF+FC

                   ⇒ c = AD+CE

                   ⇒ c = (AB-DB)(CB-EB)

                   ⇒ c = a-r +b-r

                   ⇒ 2r =a+b-c

                        r =( a+b-c)/2

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