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कि =(5) उपू- A LIRS S0l es o—b," 'h?u» Vi W.:‘M,,WJ.& PY"Ld".[’" नह ९५८४, (लि फी पएिप्प्ट-जप्छ रिप डवर्व6०- ही... s e Jj,‘,न.कि o : . |
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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 So, AE = AF and BD = BFAlso CE = CD = rand b-r = AF, a- r = BF Therefore, AB = AF + BFc = b-r + a-r Hence, r = a+b-c/2 |
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