# A cylindrical bucket, 32 cm high and with radius of base 18 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 cm, find the radius and slant height of the heap.

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Given:

A cylindrical bucket, $32 \mathrm{~cm}$ high and with radius of base $18 \mathrm{~cm}$, is filled with sand.

This bucket is emptied out 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 cylindrical bucket $r=18 \mathrm{~cm}$

Height of the cylindrical bucket $h=32 \mathrm{~cm}$

This implies,

The volume of the sand in the bucket $=\pi r^{2} h$

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

$=\pi \times 324 \times 32$

$=10368 \pi \mathrm{cm}^{3}$

Height of the conical heap $H=24 \mathrm{~cm}$ Let the radius of the conical heap be $R$.

This implies,

Volume of the conical heap $=\frac{1}{3} \pi R^{2} H$

$\Rightarrow 10368 \pi=\frac{1}{3} \times \pi R^{2} \times 24$

$\Rightarrow R^{2}=\frac{10368 \pi \times 3}{\pi \times 24}$

$=1296$

$=(36)^{2}$

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

Therefore,

The radius of the conical heap $=36 \mathrm{~cm}$

Slant height of the heap $l=\sqrt{R^{2}+H^{2}}$

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

$=\sqrt{1296+576}$

$=\sqrt{1872}$

$=\sqrt{144\times13}$

$=12\sqrt{13}\ cm$

The radius and slant height of the heap are $36\ cm$ and $12\sqrt{13}\ cm$ respectively.

Updated on 10-Oct-2022 13:25:34