Indian Institute of Foreign Trade 2010 Solved Paper

Let us assume that 'a' is the length of the side of equal sides of the isosceles triangle.We can see that $\mathit{a}+\sqrt{2}\times \mathit{a}+\mathit{a}=2$
$\Rightarrow \mathit{a}=\frac{2}{2+\sqrt{2}}$
Therefore, the perimeter of the regular octagon $=8\times \mathit{a}\times \sqrt{2}$
$\Rightarrow \frac{8\sqrt{2}\times 2}{2+\sqrt{2}}$
$\Rightarrow \frac{16}{2+\sqrt{2}}$...(1)
Area of regular octagon $=2{\mathit{x}}^2\left(1+\sqrt{2}\right)$ where 'x' is the length of each side.
Here $\mathit{x}=\mathit{a}\sqrt{2}=\frac{2\sqrt{2}}{2+\sqrt{2}}=\frac{2}{1+\sqrt{2}}$
Therefore, area of the regular octagon Hence, we can say that none of the given answer is correct.