| The area of shaded design in the given figure is (where ABCD is a square of side 10 cm and semi-circles are drawn with each of the side as diameter.) [take \[\pi =3.14\]] |
 |
A) \[57\text{ }c{{m}^{2}}\]
B) \[67c{{m}^{2}}\]
C) \[86\,c{{m}^{2}}\]
D) \[52c{{m}^{2}}\]
Correct Answer: A
Solution :
| Let the four unshaded regions be denoted by I, II, III and IV as shown in the figure. |
| Given, side of square =10 cm |
| Diameter of each semi-circle = 10 cm |
| \[\therefore \]Radius of each semi-circle \[=\frac{10}{2}=5\,cm\] |
| Now, area of I region + area of III region |
| = Area of ABCD |
| - Area of two semi-circles each of radius 5 cm |
| \[=\left( 10\times 10-2\times \frac{1}{2}\pi \times {{5}^{2}} \right)\] |
| \[=\left( 100-3.14\times 25 \right)\] |
| \[=\left( 100-78.5 \right)=21.5\,c{{m}^{2}}\] |
| Similarly, area of II region |
| + area of IV region = \[21.5\text{ }c{{m}^{2}}\] |
| \[\therefore \]Area of the shaded design |
| = Area of ABCD |
| - Area of (I + II + III + IV) region |
| \[=\left( 100-2\times 2.15 \right)=\left( 100-43 \right)=57c{{m}^{2}}\] |
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