| Refer Design II |
| Area of the remaining portion of the square ABCD is |
A) \[378\text{ }c{{m}^{2}}\]
B) \[~260\text{ }c{{m}^{2}}\]
C) \[340c{{m}^{2}}\]
D) \[278c{{m}^{2}}\]
Correct Answer: A
Solution :
Given, radius of each circle, r = 7 cm \[\therefore \]Diameter of circle, \[d=14\,\,cm\] [\[\because \]diameter = \[=2\times \]radius] In the given figure, horizontal three circles touch each other. \[\therefore \] Length of a side of square \[=3\times \]Diameter of one circle\[=\text{ }3\times 14=42\text{ }cm\] Now, area of one circle \[=\pi {{r}^{2}}=\pi {{\left( 7 \right)}^{2}}\] \[=\frac{22}{7}\times {{\left( 7 \right)}^{2}}=154\,c{{m}^{2}}\] \[\therefore \] Area of nine circles \[=9\times 154\,=1386\,c{{m}^{2}}\] Area of square ABCD \[={{\left( Side \right)}^{2}}\] \[={{\left( 42 \right)}^{2}}=1764\,c{{m}^{2}}\] Hence, area of the remaining portion of the handkerchief = Area of square Area of nine circles \[=1764-1386\text{ }=378\text{ }c{{m}^{2}}\] (iii) \[1764\,c{{m}^{2}}\] (iv) \[154\,c{{m}^{2}}\] (v) \[378\text{ }c{{m}^{2}}\]
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