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
$=2 \pi(0.7)(2.4)+\pi(0.7)^2+\pi(0.7)(2.5)$
$=3.36 \pi+0.49 \pi+1.75 \pi$
$=5.6 \pi$
$=5.6 \times \frac{22}{7}$
$=17.6 cm^2=18 cm^2\left(\text { to the nearest } cm^2\right)$
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