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 $c{m}^2$.

Answer 1

Given :

radius $\left(r\right)=\frac{diameter}{2}=\frac{1.4}{2}=0.7cm$

height $\left(h\right)=2.4cm$

Total surface area$=$ surface area of cylinder$+$ surface area of cone$-2$(surface area of above circle).

$=2\pi rh+2\pi r^2+\pi r^2+\pi r\sqrt{h^2+r^2}-2\pi r^2$

$=2\pi rh+\pi r\sqrt{h^2+r^2}+\pi r^2$

$=\pi (2rh+r^2+r\sqrt{h^2+r^2})$

$=\frac{22}{7}(2\times 0.7\times 2.4+(0.7{)}^2+0.7\sqrt{(0.7{)}^2+(24{)}^2})$

$=\frac{22}{7}(3.36+0.49+1.75)$

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

Total surface area$=17.6c{m}^2$.

Hence, this is the answer.