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 $=2.4$ cm

diameter of solid cylinder $=1.4$ cm

$l=\sqrt{r^2+h^2}=\sqrt{(0.7{)}^2+(2.4{)}^2}$

$=2.5$

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.

Surface Area

$=\left(\pi r^2\right)+\left(2\pi rh\right)+\left(\pi rl\right)$

$=\pi \left(\right(0.7{)}^2+\left(2\right)\left(0.7\right)\left(2.4\right)+\left(0.7\right)\left(2.5\right))$

$=\pi \left(5.6\right)$

$=\frac{22}{7}\left(5.6\right)$

$=17.6c{m}^2$