The radii of two circular faces of the frustum of a cone of height 21 cm are 3 cm and 2 cm respectively. What is the volume of the frustum of the cone in cm$$^3$$ ($$\pi = \frac{22}{7}$$)?

As per the question,
It is given that the height of the frustum of the cube $$(h)=21$$cm
The radius of the larger face of frustum of the cube $$(R)=3$$cm
The radius of the smaller face of frustum of the cube $$(r)=2$$cm
$$\pi = \frac{22}{7}$$
The volume of the frustum of the cube $$(V)=\dfrac{\pi h(R^2+r^2+R \times r)}{3}$$

Now substituting the value $$V=\dfrac{22\times 21 (3^2+2^2+3\times 2}{7\times3}$$
$$\Rightarrow V=\dfrac{22\times 21 (3^2+2^2+3\times 2}{7\times3}$$
$$\Rightarrow V=418 cm^3$$