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 sides of 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. $noneofthese$

Answer 1

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


Let the side of pentagon = x cm
Due to symmetry the right angled triangle cut out at corner must be isosceles right angled triangle.
$\Rightarrow x^2=y^2+y^2$
or $x=\sqrt{2}y$
As the side of square = 1 cm
$x+2y=1$
$\sqrt{2}y+2y=1\Rightarrow y=\frac{1}{\sqrt{2}(\sqrt{2}+1)}m$
Area of triangle $=\frac{1}{2}\times y\times =\frac{1}{2}{y}^2=\frac{1}{2}=\frac{1}{2(\sqrt{2}+1{)}^2}$
$=\frac{1}{4}.\frac{1}{(\sqrt{2}+1{)}^2}sq.m$
Area of 4 triangles $=\frac{1}{(\sqrt{2}+1{)}^2}$
Area of octagon = Area of square - Area of 4 triangles
$=1-\frac{1}{(\sqrt{2}+1{)}^2}=(1+\frac{1}{\sqrt{2}+1})(1-\frac{1}{\sqrt{2}+1})=\frac{\sqrt{2}}{(\sqrt{2}+1)}$
Rationalising $\to \frac{2(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)}=2(\sqrt{2}-1)$