• # question_answer A square, whose side is 2 cm, has its corners cut away so as to form a regular octagon, area of this octagon is A) $8\,(\sqrt{2}-1)$ B) $2\,(\sqrt{2}+1)$ C) $4\,(\sqrt{2}+1)$ D) none of these

 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 $4\,\,\Delta$
 $=2\times 2-4\times \frac{1}{2}\cdot {{x}^{2}}$ $=4-\frac{4\times 1}{2}\cdot {{\left[ \sqrt{2}\,(\sqrt{2}-1) \right]}^{2}}$ $=4-4\,{{(\sqrt{2}-1)}^{2}}$ $=4\left[ (1-2-1+2\sqrt{2}) \right]$ $=8\,(\sqrt{2}-1)$