 # A metallic bucket, open at the top, of height 24 cm `
Question:

A metallic bucket, open at the top, of height 24 cm is in the form of the frustum of a cone, the radii of whose lower and upper circular ends are 7 cm and 14 cm, respectively. Find
(i) the volume of water which can completely fill the bucket;
(ii) the area of the metal sheet used to make the bucket.

Solution:

We have,

Height, $h=24 \mathrm{~cm}$,

Upper base radius, $R=14 \mathrm{~cm}$ and lower base radius, $r=7 \mathrm{~cm}$

Also, the slant height, $l=\sqrt{(R-r)^{2}+h^{2}}$

$=\sqrt{(14-7)^{2}+24^{2}}$

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

$=\sqrt{49+576}$

$=\sqrt{625}$

$=25 \mathrm{~cm}$

(i) Volume of the bucket $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 24 \times\left(14^{2}+7^{2}+14 \times 7\right)$

$=\frac{22}{7} \times 8 \times(196+49+98)$

$=\frac{176}{7} \times 343$

$=8624 \mathrm{~cm}^{3}$

So, the volume of water which can completely fill the bucket is $8624 \mathrm{~cm}^{3}$.

(ii) Surface area of the bucket $=\pi(R+r) l+\pi r^{2}$

$=\frac{22}{7} \times(14+7) \times 25+\frac{22}{7} \times 7 \times 7$

$=\frac{22}{7} \times 21 \times 25+22 \times 7$

$=22 \times 3 \times 25+22 \times 7$

$=1650+154$

$=1804 \mathrm{~cm}^{2}$

So, the area of the metal sheet used to make the bucket is $1804 \mathrm{~cm}^{2}$.