Question

A tent is made in the form of a conical frustum surmounted by a cone. The diameters of the base and the top of the frustum are $20m$ and $6m$ respectively and the height is $24m$. If the height of the tent is $28m$, find the quantity of canvas required?

Answer 1

Hint: For answering this question we will use the given information and find the values of slant height of frustum and cone and use it to obtain lateral surface area because the sum of the lateral surface area is equal to the quantity of canvas required. We have formula as
$\begin{align}
  & \text{lateral surface area of frustum + lateral surface area of cone} \\
 & \Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right){{l}_{1}}+\pi {{r}_{2}}{{l}_{2}} \\
\end{align}$

Complete step-by-step solution
Now considering the question we have a tent made in the form of a conical frustum surmounted by a cone. The diameters of the base and the top of the frustum are $20m$ and $6m$ respectively and the height is $24m$ and the height of the tent is $28m$.
Let be $h$ the height of the frustum and ${{r}_{1}}$ and ${{r}_{2}}$ be the radii of the circular bases.
We have $h=24m,{{r}_{1}}=10m,{{r}_{2}}=3m$ .
From the given information we can draw a diagram as follows

Here we need to find the quantity of canvas required that is the sum of the lateral surface areas of the frustum and cone that is mathematically given as $\begin{align}
  & \text{lateral surface area of frustum + lateral surface area of cone} \\
 & \Rightarrow \pi \left( {{r}_{1}}+{{r}_{2}} \right){{l}_{1}}+\pi {{r}_{2}}{{l}_{2}} \\
\end{align}$ .
So we can say that the slant height of the frustum is ${{l}_{1}}=\sqrt{{{\left( {{r}_{1}}-{{r}_{2}} \right)}^{2}}+{{h}^{2}}}$.
By substituting the values and performing calculations we have ${{l}_{1}}=\sqrt{{{\left( 10-3 \right)}^{2}}+{{24}^{2}}}=25m$ .
For cone we have slant height as ${{l}_{2}}=\sqrt{O'B{{'}^{2}}+VO{{'}^{2}}}$ .
Now by substituting the values we will have ${{l}_{2}}=\sqrt{{{3}^{2}}+{{4}^{2}}}=5m$ .
By using these values we can obtain the quantity of canvas required as $\begin{align}
  & \pi \left( 10+3 \right)\times 25+\pi \times 3\times 5 \\
 & \Rightarrow 5\pi \left( 65+3 \right) \\
 & \Rightarrow 340\pi {{m}^{2}} \\
\end{align}$ .
Hence, we can conclude that the quantity of canvas required is $340\pi {{m}^{2}}$.

Note: While answering this question we should make a note that the lateral surface area of frustum is given by $\pi \left( {{r}_{1}}+{{r}_{2}} \right){{l}_{1}}$ . If we use $\pi \left( {{r}_{1}}-{{r}_{2}} \right){{l}_{1}}$ instead of that we will get $190\pi {{m}^{2}}$ which is completely wrong so here we should remember the formulae correctly.