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 $c{m}^2$.

Answer 1

Radius $=\frac{1.4}{2}=0.7$ cm and height $=2.4$ cm
Total surface area of structure = Curved surface area of cylinder + Area of top of cylinder + Curved surface area of cone
Curved surface area of cylinder
$2\pi rh$
$=2\pi \times 0.7\times 2.4$
$=3.36\pi c{m}^2$
Area of top:
$=\pi r^2$
$=\pi \times {0.7}^2$
$=0.49\pi c{m}^2$
Slant height of cone can be calculated as follows:
$l=\sqrt{h^2+r^2}$
$=\sqrt{{2.4}^2+{0.7}^2}$
$=\sqrt{5.76+0.49}$
$=\sqrt{6.25}=2.5cm$
Curved surface area of cone:
$=\pi rl$
$=\pi \times 0.7\times 2.5$
$=1.75\pi c{m}^2$
Hence, remaining surface area of structure
$3.36\pi +0.49\pi +1.75\pi$
$=5.6\pi =17.6c{m}^2$
$=18c{m}^2\left(approx\right)$