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


Solution:

Height $(h)$ of conical tent $=8 \mathrm{~m}$

Radius $(r)$ of base of tent $=6 \mathrm{~m}$

Slant height ( $I$ ) of tent $=\sqrt{r^{2}+h^{2}}$

$=\left(\sqrt{6^{2}+8^{2}}\right) \mathrm{m}=(\sqrt{100}) \mathrm{m}=10 \mathrm{~m}$

CSA of conical tent $=\pi r l$

$=(3.14 \times 6 \times 10) \mathrm{m}^{2}$

$=188.4 \mathrm{~m}^{2}$

Let the length of tarpaulin sheet required be $I$.

As $20 \mathrm{~cm}$ will be wasted, therefore, the effective length will be $(I-0.2 \mathrm{~m})$.

Breadth of tarpaulin $=3 \mathrm{~m}$

Area of sheet $=$ CSA of tent

$[(I-0.2 \mathrm{~m}) \times 3] \mathrm{m}=188.4 \mathrm{~m}^{2}$

$1-0.2 m=62.8 m$

$I=63 \mathrm{~m}$

Therefore, the length of the required tarpaulin sheet will be $63 \mathrm{~m}$.
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