Question

A bucket of height $8cm$ and made up of copper sheet is in the form of frustum of a right circular cone with radii of its lower and upper ends as $3cm$ and $9cm$ respectively. Calculate
(i) the height of the cone of which the bucket is a part
(ii) the volume of water which can be filled in the bucket
(iii) the area of copper sheet requried to make the bucket

Answer 1

Let $h$ be the height, $l$ the slant height and $r_1$ and $r_2$ the radii of the circular bases of a frustum of a cone

We have $h=8cm,r_1=9cm,r_2=3cm$

(i) Let $h_1$ be the height of the cone of which the bucket is a part. Then
$h_1=\frac{h{r}_1}{r_1-r_2}\Rightarrow h_1=\left(\frac{8\times 9}{9-3}\right)cm=12cm$

(ii) volume of the water which can be filled in the bucket = volume of the frustum
$=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h=\frac{1}{3}\pi (9^2+9\times 3+3^2)\times 8{cm}^3=312\pi {cm}^3$

(iii) Area of the copper sheet required to make the bucket
$=\pi (r_1-r_2)l+\pi r_2^2=\pi (9+3)\times \sqrt{{(9-3)}^2+8}+\pi \times 3^2=129\pi {cm}^2$