Question

From a solid cylinder whose height is $2.4cm$ and diameter $1.4cm$, 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 ${cm}^2$.

Answer 1

We have,

Height of solid cylinder $=$ 2.4cm

diameter of solid cylinder $=$ 1.4cm

clearly,
Total surface area of the remaining solid $=$ Curved surface area of the cylinder + Area of the base of the cylinder + Curved surface area of the cone.

$=2\pi rh+\pi r^2+\pi r\sqrt{r^2+h^2}$
$=2\times \frac{22}{7}\times 0.7\times 2.4+\frac{22}{7}\times {\left(0.7\right)}^2+\frac{22}{7}\times 0.7\times \sqrt{{\left(2.4\right)}^2+{\left(0.7\right)}^2}[\because l=\sqrt{r^2+h^2}]$

$=10.56+1.54+5.5c{m}^2$

$=17.6c{m}^2$