A) \[\frac{\pi x{{r}^{2}}}{45{}^\circ }\]
B) \[\frac{2\pi x{{r}^{2}}}{360{}^\circ }\]
C) \[\frac{2\pi x{{r}^{2}}}{45{}^\circ }\]
D) \[\frac{4\pi x{{r}^{2}}}{180{}^\circ }\]
Correct Answer: C
Solution :
Area of the sector with central angle \[\theta \] and radius r is \[\frac{\theta }{360{}^\circ }\times \pi {{r}^{2}}\] Here, radius is 4r and angle \[x{}^\circ \]. \[\therefore \] Area of sector \[=\frac{x{}^\circ }{360{}^\circ }\times \pi {{\left( 4r \right)}^{2}}\] \[=\frac{16\pi x{{r}^{2}}}{360{}^\circ }\] \[=\frac{2\pi x{{r}^{2}}}{45{}^\circ }\]
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