Question

From a solid cylinder of height $7$ cm and base diameter $12$ cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid. $[Use\pi =\frac{22}{7}].$

Answer 1

Given:

Height $\left(h\right)$ of cylindrical part $=$ height $\left(h\right)$ of the conical part $=7$ cm

Diameter of the cylindrical part $=12$ cm

Therefore, Radius $\left(r\right)$ of the cylindrical part $=\frac{12}{2}=6$ cm

So, Radius of the conical part $=6$ cm

Slant height $\left(l\right)$ of conical part =$\sqrt{r^2+h^2}$ cm

$=\sqrt{6^2+7^2}$

$=\sqrt{36+49}$

$=\sqrt{85}$

$\approx 9.22$ cm.

The total surface area of the remaining solid $=$ CSA of the cylindrical part $+$ CSA of the conical part $+$ Base area of the circular part

$=2\pi rh+\pi rl+\pi r^2$

$=2\times \frac{22}{7}\times 6\times 7+\frac{22}{7}\times 6\times 9.22+\frac{22}{7}\times 6\times 6$

$=264+173.86+113.14$

$=551$ $c{m}^2$