A 5-m-wide cloth is used to make a conical tent of base diameter 14 m and height 24 m.

Solution:

(i)

We have,

the height of the cone, $h=24 \mathrm{~m}$, the base diameter of the cone, $d=14 \mathrm{~m}$

Also, the base radius of the cone, $r=\frac{d}{2}=\frac{14}{2}=7 \mathrm{~m}$

The slant height of the cone, $l=\sqrt{h^{2}+r^{2}}$

$=\sqrt{24^{2}+7^{2}}$

$=\sqrt{576+49}$

$=\sqrt{625}$

$=25 \mathrm{~m}$

The curved surface area of the tent $=\pi r l$

$=\frac{22}{7} \times 7 \times 25$

$=550 \mathrm{~m}^{2}$

$\Rightarrow$ The area of cloth required to make the tent $=550 \mathrm{~m}^{2}$

$\Rightarrow$ The length of the cloth $=\frac{550}{5}=110 \mathrm{~m}$

So, the cost of cloth used $=110 \times 25=$ ₹ 2750

(ii)
The ratio of radius and height of a solid right-circular cone is 5:12.
Let radius, r = 5x and height, h=12x.
Volume = 314 cm3.

$\Rightarrow \frac{1}{3} \pi r^{2} h=314$

$\Rightarrow \frac{1}{3} \times 3.14 \times(5 x)^{2} \times(12 x)=314$

$\Rightarrow x^{3}=\frac{314 \times 3}{3.14 \times 5 \times 5 \times 12}$

$\Rightarrow x^{3}=1 \Rightarrow x=1$

So, radius r = 5 cm and height h = 12 cm.

Using Pythagoras Theorem, slant height is given by $l=\sqrt{r^{2}+h^{2}}=\sqrt{25+144}=\sqrt{169}=13 \mathrm{~cm}$

Total Surface Area of Cone $=\pi r(r+l)=3.14 \times 5 \times(5+13)=3.14 \times 5 \times 18=282.6 \mathrm{~cm}^{2} .$