Question

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

Answer 1

Radius of lower circular end $=r=7cm$
Radius of upper circular end $=R=14cm$
height of bucket $=h=24cm$

Slant height of frustum:

$l=\sqrt{{(R-r)}^2+h^2}$

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

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

$=\sqrt{625}$

$=25cm$

(i) Volume of frustum of cone $=\frac{1}{3}\pi h(R^2+r^2+Rr)c{m}^3$

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

$=\frac{1}{3}\times \frac{22}{7}\times 24\times (196+98+49)$

$=\frac{1}{3}\times \frac{22}{7}\times 24\times \left(343\right)$

$=8624c{m}^3$

(ii) Curved surface area $=x(R+r)$

$=22/7\times 25\times (14+7)$

$=1650c{m}^2$

Area of the base of bucket = $\pi r^2$

(consider the lower base of a bucket)

$=22/7\times 7\times 7$

$=154c{m}^2$

Area of metal sheet used to make the bucket = curved surface area + Area of the base
$=1650+154$
$=1804$
Area pf metal sheet used to make the bucket is $1804c{m}^2$