A) \[8\,(\sqrt{2}-1)\]
B) \[2\,(\sqrt{2}+1)\]
C) \[4\,(\sqrt{2}+1)\]
D) none of these
Correct Answer: A
Solution :
| Let x be the length of cutted portion than\[2-2x=\sqrt{2}\,\,x\] |
| \[x\,(2+\sqrt{2})=2\] |
| \[x=\frac{2}{\sqrt{2}\,(\sqrt{2}+1)}=\frac{\sqrt{2}}{1}\times (\sqrt{2}-1)\] |
| Area of octagon = Area of square \[-\]Area of \[4\,\,\Delta \] |
 |
| \[=2\times 2-4\times \frac{1}{2}\cdot {{x}^{2}}\] |
| \[=4-\frac{4\times 1}{2}\cdot {{\left[ \sqrt{2}\,(\sqrt{2}-1) \right]}^{2}}\] |
| \[=4-4\,{{(\sqrt{2}-1)}^{2}}\] |
| \[=4\left[ (1-2-1+2\sqrt{2}) \right]\] |
| \[=8\,(\sqrt{2}-1)\] |
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