The radii of a frustum of a cone are $14 \,cm$ and $7 \,cm$. If its height is $24 \,cm$,find its curved surface area,total surface area,and volume.

(N/A) For the given frustum of a cone,
bigger radius $r_{1} = 14 \,cm$,smaller radius $r_{2} = 7 \,cm$,and height $h = 24 \,cm$.
Slant height of the frustum of a cone $l = \sqrt{h^{2} + (r_{1} - r_{2})^{2}}$
$= \sqrt{24^{2} + (14 - 7)^{2}}$
$= \sqrt{576 + 49}$
$= \sqrt{625} = 25 \,cm$.
Curved Surface Area $(CSA)$ $= \pi l(r_{1} + r_{2})$
$= \frac{22}{7} \times 25 \times (14 + 7)$
$= \frac{22}{7} \times 25 \times 21 = 22 \times 25 \times 3 = 1650 \,cm^{2}$.
Total Surface Area $(TSA)$ $= \pi l(r_{1} + r_{2}) + \pi r_{1}^{2} + \pi r_{2}^{2}$
$= 1650 + \frac{22}{7} \times (14^{2} + 7^{2})$
$= 1650 + \frac{22}{7} \times (196 + 49)$
$= 1650 + \frac{22}{7} \times 245$
$= 1650 + 22 \times 35 = 1650 + 770 = 2420 \,cm^{2}$.
Volume $= \frac{1}{3} \pi h(r_{1}^{2} + r_{2}^{2} + r_{1}r_{2})$
$= \frac{1}{3} \times \frac{22}{7} \times 24 \times (196 + 49 + 98)$
$= \frac{22 \times 8}{7} \times 343$
$= 22 \times 8 \times 49 = 8624 \,cm^{3}$.
Thus,the $CSA$ is $1650 \,cm^{2}$,$TSA$ is $2420 \,cm^{2}$,and volume is $8624 \,cm^{3}$.