The radii of the ends of a frustum of a solid right-circular cone 45 cm high are 28 cm and 7 cm. If this frustum is melted and reconstructed into a solid right circular cylinder whose radius of base and height are in the ratio 3 : 5, find the curved surface area (in $$cm^{2}$$) of this cylinder. [Use $$\pi = \frac{22}{7}$$]

Volume of frustum = $$\dfrac{1}{3}\times\pi H\times\left(R^2+r^2+Rr\right)$$

$$\dfrac{1}{3}\times\pi\left(45\right)\times\left(28^2+7^2+\left(28\times7\right)\right)$$

$$15\pi\ \times49\times21$$

Volume of cylinder = $$\pi\ r^2h=\pi\ \left(3x\right)^2\left(5x\right)=45\pi\ x^3$$

$$45\pi\ x^3=15\pi\ \times49\times21$$

$$x=7$$

CSA of cylinder = $$2\pi\ rh=2\times\dfrac{22}{7}\times\left(3\times7\right)\times\left(5\times7\right)=4620\ cm^2$$