A bucket is in the form of a frustum of a cone with a capacity of $ 12317.6 \mathrm{~cm}^{3} $ of water. The radii of the top and bottom circular ends are $ 20 \mathrm{~cm} $ and $ 12 \mathrm{~cm} $ respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use $ \pi=3.14) . \quad $

Given:

A bucket is in the form of a frustum of a cone with a capacity of \( 12317.6 \mathrm{~cm}^{3} \) of water.

The radii of the top and bottom circular ends are \( 20 \mathrm{~cm} \) and \( 12 \mathrm{~cm} \) respectively. 

To do:

We have to find the height of the bucket and the area of the metal sheet used in its making.

Solution:

Volume of the bucket $= 12308.8\ cm^3$

Upper radius of the bucket $r_1 = 20\ cm$

Lower radius of the bucket $r_2 = 12\ cm$

Let $h$ be the height of the bucket.

Therefore,

$12308.8=\frac{\pi}{3}[r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}] \times h$

$\Rightarrow 12308.8=\frac{3.14}{3}[20^{2}+20 \times 12+12^{2}] \times h$

$\Rightarrow 12308.8=\frac{3.14}{3}[400+240+144] \times h$

$\Rightarrow \frac{12308.8 \times 3}{3.14}=784 h$

$\Rightarrow h=\frac{12308.8 \times 3}{3.14 \times 784}$

$\Rightarrow h=15\ cm$

Slant height of the bucket $l=\sqrt{(h)^{2}+(r_{1}-r_{2})^{2}}$

$=\sqrt{(15)^{2}+(20-12)^{2}}$

$=\sqrt{(15)^{2}+(8)^{2}}$

$=\sqrt{225+64}$

$=\sqrt{289}$

$=17 \mathrm{~cm}$

Surface area of the bucket $=\pi(r_{1}+r_{2}) l+\pi r_{1}^{2}$

$=\pi[(r_{1}+r_{2}) l+r_{1}^{2}]$

$=3.14[(20+12) \times 15+(12)^{2}]$

$=3.14(32 \times 15+(12)^{2}]$

$=3.14 \times(480+144]$

$=3.14 \times 624$

$=1959.36 \mathrm{~cm}^{2}$

The height of the bucket is $15\ cm$ and the area of the metal sheet used in its making is $1959.36 \mathrm{~cm}^{2}$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

125 Views

Kickstart Your Career

Get certified by completing the course

Get Started