A bucket open at the top is in the form of a frustum of a cone with a capacity of $12308.8\ cm^{3}$. The radii of the top and bottom of circular ends of the bucket are $20\ cm$ and $12\ cm$ respectively. Find the height of the bucket and also the area of the metal sheet used in making it. $( Use\ \pi = 3.14)$

Given: Volume of the Bucket$=123308.8\ cm^{3}$, Radii of the top $R=20\ cm$ and radii of the bottom of $r=12\ cm$.

To do: To find the height and area of the metal sheet used in making the frustum shaped bucket.

Solution:

Let the height of the bucket be $h\ cm$ and slant height be $l\ cm$. 

We know that volume of bucket $=\frac{\pi h}{3}.( r^{2}+R^{2}+rR)$

$\Rightarrow 123308.8=3.14\times \frac{h}{3}( 20^{2}+12^{2}+20\times12)$

$Rightarrow h=15\ cm$

Now, The slant height of the frustum, $l=\sqrt{h^{2}+( R-r)^{2}}$

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

$\Rightarrow l=\sqrt{225+64}$

$\Rightarrow l=\sqrt{289}$

$\Rightarrow l=17\ cm$

Area of metal sheet used in making it, $A=\pi r^{2}+\pi ( r+R)l$

$\Rightarrow A=3.14\times12^{2}+3.14( 20+12)17$

$\Rightarrow A=3.14\times144+3.14\times32\times17$

$\Rightarrow A=2160.32\ cm^{2}$

Hence, The height of the frustum shaped bucket $h=15\ cm$ and area of the metal sheet used to make it, $A=2160.32\ cm^{2}$.
 

Updated on: 10-Oct-2022

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