Question 50

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waecmaths question: 

An open cone with base radius 28cm and perpendicular height 96cm was stretched to form a sector to form a sector of a circle. Calculate the area of the sector. [Take $\pi =\tfrac{22}{7}$]

Option A: 

8800cm2

Option B: 

8448cm2

Option C: 

440cm2

Option D: 

4224cm2

waecmaths solution: 

      $\begin{align}  & \text{Considering the cone }PQRS \\ & l=\sqrt{{{96}^{2}}+{{28}^{2}}} \\ & l=\sqrt{9216+784} \\ & l=100cm \\ & \begin{array}{*{35}{l}}   \text{Consider the sector }ABC\text{ and the cone }PQRS,  \\   \text{The length of the }\sec \text{tor}ABC\text{ is the same the area }\!\!~\!\!\text{ of the base circle of the cone}  \\   \text{Therefore,}  \\\end{array} \\\end{align}$\[\begin{align}  & \frac{\theta }{{{360}^{{}^\circ }}}\times 2\pi l=2\pi r \\ & \frac{\theta }{{{360}^{{}^\circ }}}\times 2\pi (100)=2\pi (28) \\ & \frac{\theta }{{{360}^{{}^\circ }}}=\frac{28}{100}----(i) \\ & \text{Area of the sector }ABC \\ & A=\frac{\theta }{{{360}^{{}^\circ }}}\times \pi {{l}^{2}} \\ & \text{Substitute }\frac{28}{100}\text{ for }\frac{\theta }{{{360}^{{}^\circ }}} \\ & A=\frac{28}{100}\times \frac{22}{7}\times {{(100cm)}^{2}}=8800c{{m}^{2}} \\\end{align}\]

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