The radii of the two circular faces of the frustum of a cone are 5 cm and 4 cm.If the height of the frustum is 21 cm, what is it volume in cm$$^3$$?

($$\pi = \frac{22}{7}$$)

Volume of frustum of cone = $$\frac{1}{3}\times\ \pi\ \times\ h\times\ \left(r1^2+r2^2+\left(r1\times\ r2\right)\right)$$

                                    = $$\frac{1}{3}\times\ \frac{22}{7}\times\ 21\times\ \left(5^2+4^2+\left(5\times\ 4\right)\right)$$

                                    = $$22\times\ \left(25+16\left(5\times\ 4\right)\right)$$

                                    = $$22\times\ \left(41+20\right)$$

                                    = $$22\times\ 61=1342$$