Let $$C_1$$ and $$C_2$$ be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm area of C, and 5 cm then $$\frac{area of C_1}{area of C_2}$$

Sides of the triangle = 3, 4 and 5 cm

$$\because$$ $$3^2+4^2 = 25 = 5^2$$

Clearly, it is a right angled triangle.

=> Area of $$\triangle$$ = $$\frac{1}{2}$$ * 3 * 4 = 6 $$cm^2$$

Let radius of $$C_1$$ = $$r_1$$ and of $$C_2$$ = $$r_2$$

Using the formula, Area = inradius * semi perimeter

=> $$r_1$$ = $$\frac{6}{\frac{3+4+5}{2}}$$

=> $$r_1$$ = 1 cm

Also, area = $$\frac{a*b*c}{4*R}$$ [where a,b,c are sides of triangle and R is circumradius]

=> $$r_2$$ = $$\frac{3*4*5}{4*6}$$

=> $$r_2$$ = $$\frac{5}{2}$$ cm

To find : $$\frac{area of C_1}{area of C_2}$$

= $$\frac{\pi r_1^2}{\pi r_2^2}$$

= $$\frac{1}{\frac{25}{4}}$$ = $$\frac{4}{25}$$