The height of a conical tent is 14 m and its floor area is 346.5 m2. How much canvas, 1.1 wide, will be required for it?
(a) 490 m
(b) 525 m
(c) 665 m
(d) 860 m
(b) 525 m
Area of the floor of a conical tent $=\pi r^{2}$
Therefore,
$\pi r^{2}=346.5$
$\Rightarrow \frac{22}{7} \times r^{2}=346.5$
$\Rightarrow r^{2}=\left(\frac{3465}{10} \times \frac{7}{22}\right)$
$\Rightarrow r^{2}=\frac{441}{4}$
$\Rightarrow r^{2}=\left(\frac{21}{2}\right)^{2}$
$\Rightarrow r=\frac{21}{2} \mathrm{~m}$
Height of the cone = 14 m
Slant height of the cone, $l=\sqrt{r^{2}+h^{2}}$
$=\sqrt{\left(\frac{21}{2}\right)^{2}+(14)^{2}}$
$=\sqrt{\frac{441}{4}+196}$
$=\sqrt{\frac{1224}{4}}$
$=\frac{35}{4} \mathrm{~m}$
Area of the canvas = Curved surface area of the conical tent
$=\pi r l$
$=\left(\frac{22}{7} \times \frac{21}{2} \times \frac{35}{2}\right) \mathrm{m}^{2}$
$=577.5 \mathrm{~m}^{2}$
Length of the canvas $=\frac{\text { Area of the canvas }}{\text { Width of the canvas }}$
$=\frac{577.5}{1.1} \mathrm{~m}$
$=525 \mathrm{~m}$