To find the inradius r of a circle we can use the formula: r = Area / semi-perimeter

Now for a circle with sides a, b and c, the semi-perimeter is (a +b +c)/2 and the area is sqrt [s (s – a) (s – b) (s-c)]

Here we can find the semi-perimeter as (18 + 24 + 36)/2 = 39.

Area = sqrt [(39 – 18) (39 – 24) (39 – 36) (39)]

= sqrt [21*15*3*39]

= 9*sqrt (7*5 *13)

= 9* sqrt 455

Therefore r = 9*sqrt 455 / 39

= 3*sqrt 455/ 13

**The inradius of a triangle with sides 18, 24 and 36 centimeters is 3*sqrt 455/ 13 cm.**

The inradius r of the circle a triangle is given by:

r = Area of the triangle /2s,

r = sqrt{s(s-a)(s-b)(s-c), where 2s= a+b+c and a,b,anc are the triangle.

Given a=18, b= 24 and c= 36.

=> s = (18+24+36)/2 = 78/2 =39.

=> Area of the triangle = sqrt{39(39-18)(39-24)(39-36)} = sqrt{39*21*15*3}= 197.98

Therefore r = 197.98/39 = 4.92 nearly.

So the inradius of the given triangle = 4.92.