A) \[256\,c{{m}^{2}}\]
B) \[128\,c{{m}^{2}}\]
C) \[64\sqrt{2}\,c{{m}^{2}}\]
D) \[64\,c{{m}^{2}}\]
Correct Answer: B
Solution :
| [b] Given, radius of circle, \[r=OC=8cm\] |
| \[\therefore \] Diameter of the circle \[=AC=2\times OC\] |
| \[=2\times 8=16\,cm\] which is equal to the diagonal of a square. |
| Let a side of square be x. |
| In right angled \[\Delta ABC\] |
| \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\] |
| [By Pythagoras theorem] |
 |
| \[\Rightarrow \,\,\,\,\,\,\,\,{{(16)}^{2}}={{x}^{2}}+{{x}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,256=2{{x}^{2}}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,{{x}^{2}}=\frac{256}{2}=128\] |
| \[\therefore \] Area of the square \[={{x}^{2}}=128c{{m}^{2}}\] |
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