| In figure, two concentric circles with centre O, have radii \[21\,cm\] and \[42\,cm\]. If \[\angle \text{AOB}\ \text{=}\ \text{6}{{\text{0}}^{\operatorname{o}}}\] the area of the shaded region is: (CBSE 2017,19) |
 |
A) \[3465\,c{{m}^{2}}\]
B) \[1295\,c{{m}^{2}}\]
C) \[2565\,c{{m}^{2}}\]
D) \[3980\,c{{m}^{2}}\]
Correct Answer: A
Solution :
| [a] Let the sector AOB meets the inner circle at C and D. |
| Given, Radius of inner circle \[({{r}_{2}})=21cm.\] |
| Radius of external circle \[({{r}_{1}})=42\,cm\] and central angle \[\angle AOB,\] \[\theta =60{}^\circ \] |
| Area of region \[ABDC=\pi \times \frac{\theta }{360}\,(r_{1}^{2}-r_{2}^{2})\] |
| \[=\pi \times \frac{60}{360}\,({{42}^{2}}-{{21}^{2}})\] |
| \[=\frac{22}{7}\times \frac{1}{6}\times 63\times 21=693c{{m}^{2}}\] |
| So. radius of each quadrant \[(r)=\frac{12}{2}=6cm\] |
| and area of four quadrants \[=4\times \left( \frac{1}{4}\pi {{r}^{2}} \right)\] |
| \[=3.14\times {{6}^{2}}=113.04\,c{{m}^{2}}\] |
| \[\therefore \] Area of the shaded part |
| \[=\text{Area of rectangle}-\text{Area of four quadrants}\] |
| \[=(144-113.04)c{{m}^{2}}=30.96c{{m}^{2}}\] |
| \[\therefore \] Area of shaded region |
| \[=\pi ({{42}^{2}}-{{21}^{2}})-\text{Area of region ABDC}\] |
| \[=\frac{22}{7}\times 63\times 21-693\] |
| \[=4158-693=3465c{{m}^{2}}\] |
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