| In the given figure, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles, then the area of the shaded region is |
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A) \[154\text{ }c{{m}^{2}}\]
B) \[21c{{m}^{2}}\]
C) \[42c{{m}^{2}}\]
D) \[184c{{m}^{2}}\]
Correct Answer: C
Solution :
| Given, side of square =14 cm |
| i.e. AB=BC=CD=DA= 14 cm |
| \[\therefore \]Radius of circle \[=\frac{1}{2}\] (Side of a square) |
| \[=\frac{14}{2}=7\,cm\] |
| Area of quadrant of one circle \[=\frac{\pi {{r}^{2}}}{4}\] |
| \[=\frac{22}{7\times 4}\times {{\left( 7 \right)}^{2}}=\frac{154}{4}c{{m}^{2}}\] |
| \[\therefore \] Area of four quadrants of four circles |
| \[=4\left( \frac{154}{4} \right)=154c{{m}^{2}}\] |
| Now, area of square |
| \[=\text{ }{{\left( Side \right)}^{2}}={{\left( 14 \right)}^{2}}=196\text{ }c{{m}^{2}}\] |
| Hence, area of shaded region |
| = Area of square - Area of four quadrants |
| \[=196-154\text{ }=42\text{ }c{{m}^{2}}\] |
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