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The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find the area of the metal sheet used to make the bucket.[Use $\pi =3.14$ ]
Given: The diameter of the lower end of the given bucket $=10$ and diameter of upper ends of the bucket$=30 m$. Height of the bucket$=24 cm$.
To do: To find the area of the metal sheet used to make the frustum.
Solution: Diameter of the upper end of the bucket, $d_{1} =30\ cm$
$\therefore$ radius of the upper end of the bucket, $r_{1} =\frac{d_{1}}{2}=\frac{30}{2}=15\ cm$
Diameter of the lower end, $d_{2} =10\ cm$
$\therefore$ radius of the lower end of the bucket, $r_{2} =\frac{d_{2}}{2}=\frac{10}{2} = 5\ cm$
Height of the bucket$h=24 cm$.
As known slant height of the frustum $=\sqrt{( r_{2} -r_{1})^{2} +h^{2}}$
$=\sqrt{( 15-5)^{2} +24^{2}}$
$=\sqrt{100+576}$
$=\sqrt{676}$
$=26\ cm$
Area of the metal sheet to make the bucket$=$Area of the frustum$+$Area of the base
$=\pi ( r_{1} +r_{2}) l+\pi r^{2}_{2}$
$=\pi \left[( 15+5) 26+5^{2}\right]$
$=\pi [( 20\times 26) +25]$
$=\pi [( 20\times 26) +25]$
$=545\pi \ cm^{2}$
$=545\ \times 3.14$
$=17,11.3 \ cm^{2}$
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