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 $cm^2$.
✓
Answer
Diameter of the solid cylinder $= 1.4\ cm$
$\therefore$ Radius of the solid cylinder $= 0.7 \ cm$
$\therefore$ Radius of the base of the conical cavity $= 0.7 \ cm$
Height of the solid cylinder $= 2.4 \ cm$
$\therefore$ Height of the conical cavity $= 2.4 \ cm$
$\therefore$ Slant height of the conical cavity = $\sqrt{(0.7)^2\;+\;(2.4)^2\;}\;=\sqrt{0.49\;+5.76}=\;\sqrt{6.25}\;=\;2.5 \ cm$
$\therefore TSA$ of remaining solid
$ =2 \pi(0.7)(2.4)+\pi(0.7)^2+\pi(0.7)(2.5) $
$ =3.36 \pi+0.49 \pi+1.75 \pi$
$ =5.6 \pi$
$ =5.6 \times \frac{22}{7} $
$ =17.6 \mathrm{~cm}^2=18 \mathrm{~cm}^2 \text { (to the nearest } \mathrm{cm}^2 \text { ) }$
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