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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 $cm^2$.
Given:
From a solid cylinder of height \( 2.4 \mathrm{~cm} \) and diameter \( 1.4 \mathrm{~cm} \), a conical cavity of the same height and same diameter is hollowed out.
To do:
We have to find the total surface area of the remaining solid.
Solution:
Diameter of the solid cylinder $= 1.4\ cm$
This implies,
Radius of the cylinder $r = \frac{1.4}{2}$
$ = 0.7\ cm$
Height of the cylinder $h = 2.4\ cm$
Radius of the cone $r = 0.7\ cm$
Height of the cone $h = 2.4\ cm$
Slant height of the cone $l=\sqrt{r^{2}+h^{2}}$
$=\sqrt{(0.7)^{2}+(2.4)^{2}}$
$=\sqrt{0.49+5.76}$
$=\sqrt{6.25}$
$=2.5 \mathrm{~cm}$
Total surface area of the remaining solid $=$ Surface area of the cylinder $+$ Surface area of the cone
$=2 \pi r h+\pi r^{2}+\pi r l$
$=\pi r(2 h+r+l)$
$=\frac{22}{7} \times 0.7(2 \times 2.4+0.7+2.5)$
$=2.2(4.8+0.7+2.5)$
$=2.2(8)$
$=17.6 \mathrm{~cm}^{2}$
The total surface area of the remaining solid is $18\ cm^2$.