In $\triangle ABC$,$\angle B = 90^{\circ}$,$AB = 28 \, cm$,and $BC = 21 \, cm$. If the triangle is revolved about the side $AB$,find the curved surface area of the cone so obtained. Similarly,if the triangle $ABC$ is revolved about the side $BC$,find the curved surface area of the cone so obtained.

(N/A) $1$. When revolved about $AB$: The height $h = AB = 28 \, cm$ and the radius $r = BC = 21 \, cm$. The slant height $l = \sqrt{r^2 + h^2} = \sqrt{21^2 + 28^2} = \sqrt{441 + 784} = \sqrt{1225} = 35 \, cm$. The curved surface area $CSA = \pi r l = \frac{22}{7} \times 21 \times 35 = 22 \times 3 \times 35 = 2310 \, cm^2$.
$2$. When revolved about $BC$: The height $h = BC = 21 \, cm$ and the radius $r = AB = 28 \, cm$. The slant height $l = \sqrt{r^2 + h^2} = \sqrt{28^2 + 21^2} = 35 \, cm$. The curved surface area $CSA = \pi r l = \frac{22}{7} \times 28 \times 35 = 22 \times 4 \times 35 = 3080 \, cm^2$.