The radii of the ends of a frustum of a cone $45\, cm$ high are $28\, cm$ and $7\, cm$ (see figure). Find its volume,the curved surface area,and the total surface area (Take $\pi=\frac{22}{7}$).

(N/A) $(i)$ Volume of the frustum $= \frac{1}{3} \pi h (r_{1}^{2} + r_{2}^{2} + r_{1} r_{2})$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot [(28)^{2} + (7)^{2} + (28)(7)] \, cm^{3}$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot [784 + 49 + 196] \, cm^{3}$
$= \frac{1}{3} \cdot \frac{22}{7} \cdot 45 \cdot 1029 \, cm^{3} = 48510 \, cm^{3}$
$(ii)$ Slant height $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}} = \sqrt{(45)^{2} + (28 - 7)^{2}} \, cm$
$= \sqrt{2025 + (21)^{2}} = \sqrt{2025 + 441} = \sqrt{2466} \approx 49.66 \, cm$
Curved surface area $= \pi (r_{1} + r_{2}) l = \frac{22}{7} (28 + 7) (49.66) = \frac{22}{7} \cdot 35 \cdot 49.66 = 110 \cdot 49.66 = 5462.6 \, cm^{2}$
$(iii)$ Total surface area $= \pi (r_{1} + r_{2}) l + \pi r_{1}^{2} + \pi r_{2}^{2}$
$= 5462.6 + \frac{22}{7} (28)^{2} + \frac{22}{7} (7)^{2}$
$= 5462.6 + 2464 + 154 = 8080.6 \, cm^{2}$