Question 28

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Jambmaths question: 

A  chord of a circle subtends an angle of 120oat the centre of a circle of diameter $4\sqrt{3}cm$. Calculate the area of the major sector.

Option A: 

4πcm2

Option B: 

32πcm2

Option C: 

16πcm2

Option D: 

8πcm2

Jamb Maths Solution: 

$\begin{align}  & r=\frac{d}{2}=2\sqrt{3}cm \\ & \text{Area of sector }=\frac{\theta }{{{360}^{\text{o}}}}\times \pi {{r}^{2}} \\ & =\frac{{{120}^{\text{o}}}}{{{360}^{\text{o}}}}\times \pi {{(2\sqrt{3})}^{2}} \\ & =\frac{{{120}^{\text{o}}}}{{{360}^{\text{o}}}}\times 12\pi =4\pi c{{m}^{2}} \\ & \text{Area of circle}=\pi {{r}^{2}}= \\ & =\pi {{(2\sqrt{3})}^{2}}=12\pi c{{m}^{2}} \\ & \text{Area of the major sector }=(12\pi -4\pi )c{{m}^{2}}=8\pi c{{m}^{2}} \end{align}$

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