The radius of the circumcircle of a right angled triangle is 15 cm and the radius of its inscribed circle is 6 cm. Find the sides of the triangle.

Circumradius = 15 cm and inradius = 6 cm

Let sides be a,b,c

Since, it's a right triangle, => Circumradius lies on the mid point of hypotenuse

=> length of hypotenuse(c) = 15*2 = 30 cm

The sum of the other two sides of a right triangle is equal to twice the sum of inradius and circumradius.

=> $$(a+b) = 2(15+6)$$

=> a+b = 42 ----------Eqn(1)

Now, sum of all sides = 42+30 = 72 cm

Product of the two sides(other than hypotenuse) is equal to the product of inradius and sum of all three sides.

=> ab = 6 * 72

=> ab = 432 ---------Eqn(2)

From eqn(1) and (2)

=> a = 18 or b = 24 (or vice versa)

=> Sides are = 18,24,30