**Question:**

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

**Solution:**

(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 $=\frac{\text { Area of the canvas }}{\text { Width of the canvas }}$

$=\frac{577.5}{1.1} \mathrm{~m}$

$=525 \mathrm{~m}$

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