Question

The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a
(a+b-c)
circle. Prove that the radius r of the circle is given by r =
2

please solve this question in 2marks type answer ​

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 and AB=c.

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

b-r =AF, a-r=BF

or AB =c=AF+BF = b-r+a-r

r = a+b+c/2

Step-by-step explanation: