A tent is in the form of a cylinder of diameter $ 20 \mathrm{~m} $ and height $ 2.5 \mathrm{~m} $, surmounted by a cone of equal base and height $ 7.5 \mathrm{~m} $. Find the capacity of the tent and the cost of the canvas at $ ₹ 100 $ per square metre.

Given:

A tent is in the form of a cylinder of diameter \( 20 \mathrm{~m} \) and height \( 2.5 \mathrm{~m} \), surmounted by a cone of equal base and height \( 7.5 \mathrm{~m} \). 

To do:

We have to find the capacity of the tent and the cost of the canvas at \( ₹ 100 \) per square metre.

Solution:

Diameter of the base of the tent $= 20\ m$

This implies,

Radius of the tent $r =\frac{20}{2}$

$= 10\ m$

Height of the cylindrical part $h_1 = 2.5\ m$

Height of the conical part $h_2 = 7.5\ m$

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

$=\sqrt{(10)^{2}+(7.5)^{2}}$

$=\sqrt{100+56.25}$

$=\sqrt{156.25}$

$=12.5 \mathrm{~m}$

Therefore,

Capacity of the tent $=\pi r^{2} h_{1}+\frac{1}{3} \pi r^{2} h_{2}$

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

$=\pi \times 10+10(2.5+\frac{1}{3} \times 7.5)$

$=100 \pi(2.5+2.5)$

$=100 \pi \times 5$

$=500 \pi \mathrm{m}^{3}$

Total curved surface area of the tent $=$ Curved surface area of the cylindrical part $+$ Curved surface area of the conical part

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

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

$=\frac{22}{7} \times 10(2 \times 2.5+12.5)$

$=\frac{220}{7} \times(5+12.5)$

$=\frac{220}{7} \times 17.5$

$=220 \times 2.5$

$=550 \mathrm{~m}^{2}$

Cost of the canvas per $m^2=Rs.\ 100$

Therefore,

Total cost of the canvas $= Rs.\ 550 \times 100$

$=Rs.\ 55000$

The capacity of the tent is $500 \pi\ m^3$ and the cost of the canvas is Rs. 55,000.

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Updated on: 10-Oct-2022

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