A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is $ 3.5 \mathrm{~cm} $ and the heights of the cylindrical and conical portions are $ 10 \mathrm{~cm} $. and $ 6 \mathrm{~cm} $, respectively. Find the total surface area of the solid. (Use $ \pi=22 / 7 $ )

Given:

A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end.

The radius of the common base is \( 3.5 \mathrm{~cm} \) and the heights of the cylindrical and conical portions are \( 10 \mathrm{~cm} \) and \( 6 \mathrm{~cm} \) respectively.

To do:

We have to find the total surface area of the solid.

Solution:

Radius of the common base $r = 3.5\ m$
Height of the cylindrical part $h_1 = 10\ cm$
Height of the conical part $h_2 = 6\ cm$

Slant height of the conical part $l=\sqrt{r^{2}+h_{2}^{2}}$

$=\sqrt{(3.5)^{2}+(6)^{2}}$

$=\sqrt{12.25+36}$

$=\sqrt{48.25}$

$=6.95 \mathrm{~cm}$

Total surface area of the solid $=$ Curved surface area of the conical part $+$ Curved surface area of the cylindrical part $+$ Curved surface area of the hemispherical part

$=\pi r l + 2 \pi r h_{1}+2 \pi r^{2}$

$=\pi r(l+2 h_{1}+2 r)$

$=\frac{22}{7} \times 3.5(6.95+2 \times10+2 \times 3.5)$

$=11(6.95+20+7)$

$=11 \times 33.95$

$=373.45 \mathrm{~cm}^{2}$

The total surface area of the solid is $373.45\ cm^2$.

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Updated on: 10-Oct-2022

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