Question

A piece of paper is in the shape of a square of side 1 m long. It is cut at the four corners to make a regular polygon of eight sides (octagon). The area of the polygon is

• A. $2(\sqrt{2}-1)m^2$
• B. $(\sqrt{2}-1)m^2$
• C. $\frac{1}{\sqrt{2}}{m}^2$
• D. none of these

Answer 1

The correct option is A $2(\sqrt{2}-1)m^2$

let the side length of cut triangle (other than hypotenuse) be $x$.

Then the area of cut triangle (since right angled triangle) will be $\frac{x^2}{2}$.

So area of all the cut portion is four times the triangle area i.e., $2x^2$.

We know that $1-2x=\sqrt{2}x$ $\Rightarrow$ $x=\frac{1}{(2+\sqrt{2})}$.

Hence the area of octagon is $1-2x^2$=$2(\sqrt{2}-1)$