Question

What length of tarpaulin $3\mathrm{m}$ wide will be required to make conical tent of height $8\mathrm{m}$ and base radius 6m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately $20\mathrm{cm}$. [Use $\pi =3.14$]

Answer 1

Find the length of the required tarpaulin sheet .

The height of the conical tent $\left(h\right)$ is$8\mathrm{m}$.

The radius of the base of tent $\left(r\right)$ is $6\mathrm{m}$.

The Slant height of the tent is :
$$\begin{gathered} I^2=r^2+h^2 \\ \begin{array}{rcl}& \Rightarrow & I^2=6^2+8^2\\ & \Rightarrow & I^2=36+64\\ & \Rightarrow & I^2=100\\ & \Rightarrow & I=10\end{array} \end{gathered}$$
The Curved surface area is$\pi rl$.

Substitute the values in the above formulae
$$\begin{gathered} \mathrm{CSA}\mathrm{of}\mathrm{cone}=\mathrm{\pi rl} \\ =(3.14\times 6\times 10){\mathrm{m}}^2 \\ =188.4{\mathrm{m}}^2 \end{gathered}$$
Let $\mathrm{L}$ be the desired tarpaulin sheet length.
Since, $20\mathrm{cm}$ will be wasted,
Thus, the Effective length will be $(\mathrm{L}-0.2)\mathrm{m}$.
Given that the breadth of tarpaulin is $3\mathrm{m}$.
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{sheet}=\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{tent} \\ \begin{array}{rcl}\left[\right(L-0.2)\times 3]& =& 188.4\\ & \Rightarrow & L-0.2=62.8\\ & \Rightarrow & L=63\mathrm{m}\end{array} \end{gathered}$$

Hence, the required tarpaulin sheet will be $63\mathrm{m}$ long.