Find the area of the shaded region in the given figure, if radii of the two concentric circles with centre $O$ are $7\ cm$ and $14\ cm$ respectively and $\angle AOC = 40^o$
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Given:
Two concentric circles of radii $7\ cm$ and $14\ cm$ where $\angle AOC=40^{o}$ in the given figure.
To do :
We have to find the area of the shaded region, enclosed between two concentric circles
Solution:
Area of the region ABDC $=$Area of sector AOC $–$ Area of sector BOD
$=\frac{\theta }{360^{o}} \times \pi r^{2}_{1} -\frac{\theta }{360^{o}} \times \pi r^{2}_{2}$
$=\frac{40}{360^{o}} \times \frac{22}{7} \times ( 14)^{2} -\frac{40^{o}}{360^{o}} \times \frac{22}{7} \times ( 7)^{2}$
$=\frac{22}{7} \times \frac{1}{9}\times( 196-49)$
$=\frac{22\times147}{7\times9}$
$=\frac{154}{3}$
$=51.33cm^{2}$
Area of circular ring $=$Area of the outer ring$-$Area of the inner ring
$=\pi r^{2}_{1} -\pi r^{2}_{2}$
$=\frac{22}{7}( 14^{2} -7^{2})$
$=\frac{22}{7}(14+7)(14-7)$
$=22 \times (21)$
$=22\ \times \ 21$
$=462\ cm^{2}$
$\therefore$ Required shaded region $=$Area of circular ring$–$Area of region ABDC
$=462 – 51.33$
$=410.67\ cm^{2}$
Therefore, the area of shaded region is $410.67\ cm^{2}$.
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