Question 29

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Jambmaths question: 

$\begin{align}  & \text{A chord of a circle suntends an angle of 12}{{\text{0}}^{\circ }}\text{ at the centre of a circle of diameter }4\sqrt{3}cm\text{. } \\ & \text{Calculate the area of the major sector} \\ & \text{(A) }4\pi \text{  (B) }8\pi \text{  (C) }16\pi \text{  }(D)\text{ }32\pi  \\\end{align}$

Jamb Maths Solution: 

$\begin{align}  & \text{Diameter}=d=4\sqrt{3}cm \\ & \text{radius }=r=\frac{d}{2}=\frac{4\sqrt{3}}{2}cm=2\sqrt{3} \\ & \text{Area of circle}=\pi {{r}^{2}} \\ & A=\pi {{(2\sqrt{3})}^{2}}=12\pi cm \\ & \text{Are of the minor sector }=\frac{120}{360}\times \pi {{(2\sqrt{3})}^{2}}=4\pi c{{m}^{2}} \\ & \text{Area of the major sector }=(12\pi -4\pi )c{{m}^{2}}=8\pi cm \\ & \mathbf{Alternative}\text{ }\mathbf{method} \\ & r=\frac{d}{2}=2\sqrt{3}cm \\ & \text{Area of major sector}=\frac{{{360}^{\circ }}-{{120}^{\circ }}}{{{360}^{\circ }}}\times \pi {{r}^{2}}=\frac{{{240}^{\circ }}}{{{360}^{\circ }}}\times \pi {{(2\sqrt{3})}^{2}}=8\pi c{{m}^{2}} \\\end{align}$

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