A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius $ 2.5 \mathrm{~m} $ and height $ 21 \mathrm{~m} $ and the cone has the slant height $ 8 \mathrm{~m} $. Calculate the total surface area and the volume of the rocket.

Given:

A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top.

The cylinder is of radius \( 2.5 \mathrm{~m} \) and height \( 21 \mathrm{~m} \) and the cone has the slant height \( 8 \mathrm{~m} \).

To do:

We have to find the total surface area and the volume of the rocket.

Solution:

Radius of the base of the rocket $r = 2.5\ m$

Height of the cylindrical part $h_1 = 21\ m$
Slant height of the conical part $l = 8\ m$
Let the  height of the conical part be $h$.

Therefore,

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

$\Rightarrow(8)^{2}=(2.5)^{2}+h^{2}$

$\Rightarrow 64=6.25+h^{2}$

$\Rightarrow h^{2}=64-6.25$

$\Rightarrow h^{2}=57.75$

$\Rightarrow h=\sqrt{57.75}$

$\Rightarrow h=7.6$

Total surface area of the rocket $=\pi r l+2 \pi r h_{1}+\pi r^{2}$

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

$=\frac{22}{7} \times 2.5(8+2 \times 21+2.5)$

$=\frac{55}{7} \times(10.5+42)$

$=\frac{55}{7} \times 52.5$

$=412.5 \mathrm{~m}^{2}$

Volume of the rocket $=\frac{1}{3} \pi r^{2} h+\pi r^{2} h_{1}$

$=\pi r^{2}(\frac{h}{3}+h_{1})$

$=\frac{22}{7} \times(2.5)^{2}(\frac{7.6}{3}+21)$

$=\frac{22 \times 6.25}{7}(\frac{7.6+63}{3})$

$=\frac{22 \times 6.25 \times 70.6}{7 \times 3}$

$=\frac{9707.5}{21}$

$=462.26 \mathrm{~m}^{3}$

The total surface area and the volume of the rocket are $412.5\ m^2$ and $462.26\ m^3$ respectively.

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Updated on: 10-Oct-2022

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