A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be $ 13 \mathrm{~m} $ and $ 7 \mathrm{~m} $, the height of the frustum be $ 8 \mathrm{~m} $ and the slant height of the conical cap be $ 12 \mathrm{~m} $, find the canvas required for the tent. $ \quad $ (Take: $ \pi=22 / 7) $

Given:

A tent consists of a frustum of a cone capped by a cone.

The radii of the ends of the frustum are \( 13 \mathrm{~m} \) and \( 7 \mathrm{~m} \), the height of the frustum is \( 8 \mathrm{~m} \) and the slant height of the conical cap is \( 12 \mathrm{~m} \).

To do:

We have to find the canvas required for the tent.

Solution:

Radius of the bottom of the tent $r_1 = 13\ m$

Radius of the top of the tent $r_2 = 7\ m$

Height of the frustum part $h_1 = 8\ m$

Slant height of the conical cap $l_2 = 12\ m$

Let $l_1$ be the slant height of the frustum part.

Therefore,

$l_{1}=\sqrt{(h)^{2}+(r_{1}-r_{2})^{2}}$

$=\sqrt{(8)^{2}+(13-7)^{2}}$

$=\sqrt{(8)^{2}+(6)^{2}}$

$=\sqrt{64+36}$

$=\sqrt{100}$

$=10 \mathrm{~m}$

The canvas required for the tent $=$ Surface area of the whole tent

$=\pi(r_{1}+r_{2}) l_{1}+\pi r_{2} l_{2}$

$=\pi[(13+7) \times 10]+\pi(7)(12)$

$=\pi[20 \times 10]+\pi(84)$

$=200 \pi+84 \pi$

$=284 \pi$

$=284 \times \frac{22}{7}$

$=\frac{6248}{7}$

$=892.57 \mathrm{~m}^{2}$

$892.57 \mathrm{~m}^{2}$ of canvas required for the tent.

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Updated on: 10-Oct-2022

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