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The circumference of one end of a frustum of a right circular cone is 48cm and of the other end is 34cm. if the height of frustum is 11cm, it’s volume ( in cubic centimeters) is.

Last updated date: 23rd Feb 2024
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IVSAT 2024
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Hint: Here we are given with the circumference of top and bottom end of the frustum. For finding the volume we need the height and radius of the ends of the frustum. So with the help of circumference we will find the radius and then putting it in the formula will lead our solution towards the answer.

Formula used:
1. Circumference of circle = \[2\pi r\]
2. Volume of frustum = \[\dfrac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)\]

Step by step solution:
Let the radius of top end be r and of bottom end be R. the height of frustum is h.
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Now consider that the circumference of the top end is 34cm. thus we can write,
\[2\pi r = 34\]
\[ \Rightarrow r = \dfrac{{17}}{\pi }\]
Similarly let the bottom end with circumference 48cm. thus we can write,
\[2\pi R = 48\]
 \[ \Rightarrow R = \dfrac{{24}}{\pi }\]
Now we have measurements of the radius then let’s use the formula.
Volume of frustum = \[\dfrac{{\pi h}}{3}\left( {{R^2} + {r^2} + Rr} \right)\]
\[{V_{frustum}} \Rightarrow \dfrac{{\pi 11}}{3}\left( {{{\left( {\dfrac{{24}}{\pi }} \right)}^2} + {{\left( {\dfrac{{17}}{\pi }} \right)}^2} + \dfrac{{24}}{\pi } \times \dfrac{{17}}{\pi }} \right)\]
\[{V_{frustum}} \Rightarrow \dfrac{{\pi 11}}{3}\left( {\dfrac{{576}}{{{\pi ^2}}} + \dfrac{{289}}{{{\pi ^2}}} + \dfrac{{408}}{{{\pi ^2}}}} \right)\]
Adding the terms,
\[{V_{frustum}} \Rightarrow \dfrac{{\pi 11}}{3}\left( {\dfrac{{1273}}{{{\pi ^2}}}} \right)\]
On simplifying the ratios we get,
\[{V_{frustum}} \Rightarrow \dfrac{{11 \times 1273}}{{3\pi }}\]
\[{V_{frustum}} \Rightarrow \dfrac{{11 \times 1273}}{{3 \times \left( {\dfrac{{22}}{7}} \right)}}\]
On rearranging the terms
\[{V_{frustum}} \Rightarrow \dfrac{{11 \times 1273 \times 7}}{{3 \times 22}}\]
Now on calculating the ratios and making necessary cancellations,
\[{V_{frustum}} \Rightarrow \dfrac{{1273 \times 7}}{{3 \times 2}}\]
\[{V_{frustum}} \Rightarrow \dfrac{{8911}}{6}\]
\[{V_{frustum}} \Rightarrow 1485.1666\]
Considering the recurring digits,
\[{V_{frustum}} \Rightarrow 1485.1\bar 6c{m^3}\]

This is the volume of the frustum \[{V_{frustum}} \Rightarrow 1485.1\bar 6c{m^3}\]

Note that we are not directly given with radii of the ends but circumference is the hint to lead with. Also while calculating the numbers be careful! Don’t forget to write the unit of volume.
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