Question

# A sector of a circle of radius $12\mathrm{cm}$ has an angle $120°$ . By coinciding its straight edges a cone is formed. Find the volume of cone.

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Solution

## Step 1: Find the radius and height of the cone.Radius of circle, $r=12\mathrm{cm}$.Angle of sector, $\theta =120°$The length of an arc is given as,$l=\frac{\theta }{360°}×2\mathrm{\pi }r\phantom{\rule{0ex}{0ex}}=\frac{120°}{360°}×2\mathrm{\pi }×12\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi }$Circumference, of the circle, is given by:$C=2\mathrm{\pi }R$Where $R$ is the radius of the base of the cone formed.Length of an arc is the circumference of the base circle. So,$2\mathrm{\pi }R=8\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒\mathrm{R}=4$As, slant height of the cone is equal to the radius of the sector used form cone $\mathrm{Slant}\mathrm{Height}\left(l\right)=r=12$Also, in a cone$\begin{array}{rcl}\mathrm{Slant}{\mathrm{height}}^{2}& =& \mathrm{radius}\mathrm{of}\mathrm{base}\mathrm{of}{\mathrm{cone}}^{2}+\mathrm{height}\mathrm{of}{\mathrm{cone}}^{2}\\ {l}^{2}& =& {R}^{2}+{h}^{2}\\ & ⇒& {12}^{2}={4}^{2}+{h}^{2}\\ & ⇒& 144=16+{h}^{2}\\ & ⇒& 144-16={h}^{2}\\ & ⇒& h=\sqrt{128}\\ \therefore h& =& 11.28\mathrm{cm}\end{array}$Step 2: Find the volume of cone.Volume of cone:$V=\frac{1}{3}\mathrm{\pi }{R}^{\mathit{2}}h\phantom{\rule{0ex}{0ex}}\mathit{}\mathit{}\mathit{}=\frac{\mathit{1}}{\mathit{3}}\mathrm{\pi }×{4}^{\mathit{2}}×11.28\phantom{\rule{0ex}{0ex}}=\frac{\mathit{1}}{\mathit{3}}\mathrm{\pi }×16×11.28\phantom{\rule{0ex}{0ex}}=189.4{\mathrm{cm}}^{3}$Hence, the volume of the cone is $189.4{\mathrm{cm}}^{3}$.

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