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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 $ ).
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 $63\ m$.
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