From a solid cylinder of height $ 2.8 \mathrm{~cm} $ and diameter $ 4.2 \mathrm{~cm} $, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Take $ \pi=22 / 7 $ )

Given:

From a solid cylinder of height \( 2.8 \mathrm{~cm} \) and diameter \( 4.2 \mathrm{~cm} \), 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 $= 4.2\ cm$

This implies,

Radius of the cylinder $r = \frac{4.2}{2}$

$ = 2.1\ cm$

Height of the cylinder $h = 2.8\ cm$

Radius of the cone $r = 2.1\ cm$

Height of the cone $h = 2.8\ cm$

Slant height of the cone $l=\sqrt{r^{2}+h^{2}}$

$=\sqrt{(2.1)^{2}+(2.8)^{2}}$

$=\sqrt{4.41+7.84}$

$=\sqrt{12.25}$

$=3.5 \mathrm{~cm}$

Total surface area of the remaining solid $=$ Surface area of the cylinder $+$ Surface area of the cone

$=2 \pi r h+\pi r^{2}+\pi r l$

$=\pi r(2 h+r+l)$

$=\frac{22}{7} \times 2.1(2 \times 2.8+2.1+3.5)$

$=6.6(5.6+2.1+3.5)$

$=6.6(11.2)$

$=73.92 \mathrm{~cm}^{2}$

The total surface area of the remaining solid is $73.92\ cm^2$.

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Updated on: 10-Oct-2022

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