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 $cm^2$.

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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 $18\ cm^2$.

Updated on 10-Oct-2022 13:24:36