The radii of a frustum of a cone are $35 \, cm$ and $21 \, cm$ and its height is $48 \, cm$. Find the curved surface area,the total surface area,and the volume of the frustum of a cone.

(N/A) Given: $R = 35 \, cm$,$r = 21 \, cm$,$h = 48 \, cm$.
First,calculate the slant height $l = \sqrt{h^2 + (R - r)^2} = \sqrt{48^2 + (35 - 21)^2} = \sqrt{2304 + 196} = \sqrt{2500} = 50 \, cm$.
Curved Surface Area $(CSA)$ $= \pi(R + r)l = \frac{22}{7} \times (35 + 21) \times 50 = \frac{22}{7} \times 56 \times 50 = 22 \times 8 \times 50 = 8800 \, cm^2$.
Total Surface Area $(TSA)$ $= \pi l(R + r) + \pi R^2 + \pi r^2 = 8800 + \frac{22}{7} \times (35^2 + 21^2) = 8800 + \frac{22}{7} \times (1225 + 441) = 8800 + \frac{22}{7} \times 1666 = 8800 + 22 \times 238 = 8800 + 5236 = 14036 \, cm^2$.
Volume $V = \frac{1}{3} \pi h (R^2 + r^2 + Rr) = \frac{1}{3} \times \frac{22}{7} \times 48 \times (35^2 + 21^2 + 35 \times 21) = \frac{22 \times 16}{7} \times (1225 + 441 + 735) = \frac{352}{7} \times 2401 = 352 \times 343 = 120736 \, cm^3$.