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.
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.
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