A tent of height $ 77 \mathrm{dm} $ is in the form of a right circular cylinder of diameter $ 36 \mathrm{~m} $ and height $ 44 \mathrm{dm} $ surmounted by a right circular cone. Find the cost of the canvas at $ ₹ 3.50 $ per $ \mathrm{m}^{2} $ (Use $ \pi=22 / 7 $ ).

Given:

A tent of height \( 77 \mathrm{dm} \) is in the form a right circular cylinder of diameter \( 36 \mathrm{~m} \) and height \( 44 \mathrm{dm} \) surmounted by a right circular cone.

To do:

We have to find the cost of the canvas at \( ₹ 3.50 \) per \( \mathrm{m}^{2} .\)

Solution:

Total height of the tent $H= 77\ dm=7.7\ m$
Height of the cylindrical part $h_1 = 44\ dm = 4.4\ m$
Height of the conical part $h_2 = 7.7 - 4.4 = 3.3\ m$

Diameter of the base of the tent $= 36\ m$

This implies,

Radius of the base $r=\frac{36}{2}$

$=18 \mathrm{~m}$

Therefore,

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

$=\sqrt{(18)^{2}+(3.3)^{2}}$

$=\sqrt{324+10.89}$

$=\sqrt{334.89}$

$=18.3 \mathrm{~m}$

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

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

$=2 \times \frac{22}{7} \times 18 \times 4.4+\frac{22}{7} \times 18 \times 18.3$

$=\frac{22}{7} \times 18(2 \times 4.4+18.3)$

$=\frac{396}{7}(8.8+18.3)$

$=\frac{396}{7} \times 27.1$

$=1533.085 \mathrm{~m}^{2}$ Rate of canvas per $m^2=Rs.\ 3.50$

Total cost of the canvas $=Rs.\ 1533.085 \times 3.50$

$=5365.7975$

$=Rs.\ 5365.80$

The total cost of the canvas is Rs. 5365.80. 

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Updated on: 10-Oct-2022

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