If a,b,c are sides of a right triangle where c is the hypotenuse.Prove that radius r that touches the sides of triangle is r=a+b-c/2

Asked by Tarun Rath | 20th Feb, 2014, 09:36: AM

Expert Answer:

Consider the following figure.
 
Here, O is the incentre of the triangle. Thus, 'r' is the inradius of the circle.
 
a r open parentheses triangle A B C close parentheses equals a r open parentheses triangle O C B close parentheses plus a r open parentheses triangle A B O close parentheses plus a r open parentheses triangle A C O close parentheses rightwards double arrow 1 half a b equals 1 half b r plus 1 half c r plus 1 half a r rightwards double arrow a b equals r open parentheses a plus b plus c close parentheses rightwards double arrow r equals fraction numerator a b over denominator a plus b plus c end fraction
Multiply the numerator and the denominator by (a+b - c), we have,
 
r equals fraction numerator a b over denominator a plus b plus c end fraction cross times fraction numerator a plus b minus c over denominator a plus b minus c end fraction rightwards double arrow r equals fraction numerator a b open parentheses a plus b minus c close parentheses over denominator open parentheses a plus b close parentheses squared minus c squared end fraction rightwards double arrow r equals fraction numerator a b open parentheses a plus b minus c close parentheses over denominator open parentheses a plus b close parentheses squared minus a squared minus b squared end fraction rightwards double arrow r equals fraction numerator a b open parentheses a plus b minus c close parentheses over denominator 2 a b end fraction rightwards double arrow r equals fraction numerator a plus b minus c over denominator 2 end fraction

Answered by  | 20th Feb, 2014, 10:47: AM

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