Question

prove that radius of incircle of right triangle with sides a b c where c is the hypotenuse is given by a+b-c/2

Answer 1

Answer:

Step-by-step explanation:

Answer 2

1.Radius of incircle of right triangle with sides a b c where c is the hypotenuse is given by a+b-c/2

incircle means a circle which touches all sides of a triangle

and its radius is called inradius

2.I=A/S

I=inradius

A=Area of triangle

S=semiperimeter

3.Let r be radius of incircle

let a,b,c be sides of right angled triangle

r=(1/2ab)/a+b+c/2

=ab/a+b+c

4.if we consider a=3,b=4,c=5(pythagorean triplets)

we see R=3+4-5/2=3*4/3+4+5

hence proved that radius of incircle =a+b-c/2