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An open metallic bucket is in the shape of a frustum of a cone. If the diameters of the two circular ends of the bucket are $45\; cm$ and $25\; cm$ and the vertical height of the bucket is $24\; cm$ , find the area of the metallic sheet used to make the bucket. Also, find the volume of the water it can hold. (Use $\pi =\frac{22}{7}$ )

Answers (1)

Given - $R=\frac{45}{2}=22.5 \; cm$

$r=\frac{25}{2}=12.5 \; cm$

$h=24 \; cm$

The slant height of the bucket can be calculated as

$l=\sqrt{h^{2}+(R-r)^{2}}$

$l=\sqrt{24^{2}+(22.5-12.5)^{2}}$

$l=\sqrt{576+100}$

$l=26\; cm$

Metal short needed is equal to the surface of frustum and base area of the frustum.

Area of metal sheet $=\pi (R+r)l+\pi r^{2}$

$=\pi \left [ (22.5+12.5)26\right+(12.5)^{2} ]$

$=\frac{22}{7}\left [ 26\times 35+156.25 \right ]$

$=\frac{22}{7}\times 1066.2$

$=3350.9\; cm^{2}$

The volume of water = volume of the frustum

$=\frac{\pi }{3}h(R^{2}+r^{2}+Rr)$

$=\frac{22}{7\times 3}\times 24(22.5^{2}+12.5^{2}+22.5\times 12.5)$

$=23728\; cm^{3}$

Posted by

Safeer PP

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