# 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 $\pi=3.14$ ).

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Given:

The height of the conical tent is $8\ m$ and the radius of the base is $6\ m$.

The width of the tarpaulin used is $3\ m$.

To do:

We have to find the length of the tarpaulin required.

Solution:

Height of the conical tent $(h) = 8\ m$

Radius of the base $(r) = 6\ m$

Therefore,

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

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

$=\sqrt{36+64}$

$=\sqrt{100}$

$=10 \mathrm{~m}$

The curved surface area of the tent $= \pi rl$

$= 3.14 \times 6 \times 10$

$= 188.4\ m$

The width of the tarpaulin used $= 3\ m$

This implies,

Length of the tarpaulin used $= \frac{188.4}{3}$

$= 62.8\ m$

Extra length required $= 20\ cm$

$= 0.2\ m$

Total length of tarpaulin required $= 62.8 + 0.2$

$= 63\ m$

Therefore,

The total length of tarpaulin required to make a conical tent of height $8\ m$ and base radius $6\ m$ is $63\ m$.

Updated on 10-Oct-2022 13:46:36