Math, asked by anushree70, 1 year ago

pls solve it. as soon as possible.

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Answered by pavanmeena16200366

Answer:

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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