Question

The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a

circle. Prove that the radius r of the circle is given by r =(a+b+c)/2

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Answer 1

Answer:

Let the circle touches the sides BC,CA,AB of the right triangle ABC at

D,E and F respectively. Where,

BC=a

CA=b

AB=c.

Then, AE=AF and BD=BF. Also CE=CD=r.

b-r =AF, a-r=BF

r = (a+b+c)/2

Hence proved.

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Answer 2

Answer:

r=a+b-c/2

Step-by-step explanation: