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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