pls solve it. as soon as possible.
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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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pavanmeena16200366:
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