A cylindrical bucket of height $ 32 \mathrm{~cm} $ and base radius $ 18 \mathrm{~cm} $ is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is $ 24 \mathrm{~cm} $, find the radius and slant height of the heap.

Given:

A cylindrical bucket of height \( 32 \mathrm{~cm} \) and base radius \( 18 \mathrm{~cm} \) is filled with sand.

This bucket is emptied on the ground and a conical heap of sand is formed.

The height of the conical heap is \( 24 \mathrm{~cm} \).

To do:

We have to find the radius and slant height of the heap.

Solution:

Radius of the base of the bucket $= 18\ cm$

Height of the bucket $= 32\ cm$

This implies,

Volume of the sand in the cylindrical bucket $= \pi r^2 h$

$= \pi (18)^2 \times 32$

$= 10368 \pi$

Height of the conical heap of sand $h = 24\ cm$

Let the radius of the heap be $r$.
Therefore,

Volume of the sand in the heap $=\frac{1}{3} \pi r^2 h$

$=\frac{1}{3} \pi r^{2} \times 24$

$=8 \pi r^{2}$

According to the question,

Volume of the sand in the cylindrical bucket $=$ Volume of the sand in the conical heap

$\Rightarrow 10368 \pi=8 \pi r^{2}$

$\Rightarrow 10368=8 r^{2}$

$\Rightarrow r^{2}=\frac{10368}{8}$

$\Rightarrow r^{2}=1296$

$\Rightarrow r=36 \mathrm{~cm}$

Slant height of the conical heap $l=\sqrt{h^{2}+r^{2}}$

$=\sqrt{(24)^{2}+(36)^{2}}$

$=\sqrt{576+1296}$

$=\sqrt{1872}$

$=43.26 \mathrm{~cm}$

The radius and slant height of the heap are $36\ cm$ and $43.26\ cm$ respectively.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

17 Views

Kickstart Your Career

Get certified by completing the course

Get Started