Question

From a solid cylinder whose height is $2.4$ cm and diameter $1.4$ cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $c{m}^2$. (Use $\pi =\frac{22}{7}$).

Answer 1

Given:
Height (h) of the conical part = Height (h) of the cylindrical part $=2.4$ cm
Diameter of the cylindrical part $=1.4$ cm

Radius = $\frac{Diameter}{2}$
Radius(r) of the cylindrical part $=0.7$ cm

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

$=\sqrt{{0.7}^2+{2.4}^2}=\sqrt{0.49+5.76}=\sqrt{6.25}=2.5$

Total surface area of the remaining solid = CSA of cylindrical part + CSA of conical part + Area of cylindrical base

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

$=2\times \frac{22}{7}\times 0.7\times 2.4+\frac{22}{7}\times 0.7\times 2.5+\frac{22}{7}\times 0.7\times 0.7$

$=4.4\times 2.4+2.2\times 2.+2.2\times 0.7$

$=10.56+5.50+1.54=17.60$ cm${}^2$

The total surface area of the remaining solid to the nearest $c{m}^2$ is $18$ $c{m}^2$