A Sector Of A Circle Of Radius 12 Cm Has An Angle 120 . By Coinciding Its Straight Edges A Cone Is Formed. Find The Volume Of Cone.

Given the radius of the circle 12cm

This becomes the slant height of the cone

The angle of sector \(120^{\circ} \)

The length of arc is given as = \( \frac{x}{360} * 2 \Pi r \) \(\Rightarrow \frac{120}{360} * 2 * \Pi * 12 \) \(\Rightarrow \frac{1}{3}2\Pi * 12 \) \(\Rightarrow 2\Pi * 4 \) \(\Rightarrow 8\Pi \)

This become the circumference of base circle = \(2 \Pi r = 8\Pi \)

2r = 8

r = 4

According to Pythagoras theorem

\((Slantheight)^{2} = (height)^{2} + (radius)^{2} \) \(\Rightarrow 12^{2} = (height)^{2} + (4)^{2} \) \(\Rightarrow 144 = (height)^{2} + 16 \) \(\Rightarrow (height)^{2} = 144 – 16 \) \(\Rightarrow (height)^{2} = 132 \) \(\Rightarrow height = \sqrt{132} \)

The volume of cone =\( \frac{1}{3}\Pi r^{2}h \) \(\Rightarrow \frac{1}{3} * \frac{22}{7} * 4 * 4 * \sqrt{132} \) \(\Rightarrow \frac{362\sqrt{132}}{21} \)

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