# A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon?

A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon? Source: ISI BMATH UGA 2017 paper

A regular polygon with 8 sides can be divided into eight congruent triangles .I tried to find the area of a triangle in the following method. An angle of a triangle is 360/8=45..then I draw a perpendicular bisector of the angle which is height of the triangle and I found it to be 1/2 as the square is of unit length .Now I got a right angled triangle from which I wanted to find the length of the base of the triangle but I couldn't do so.

• I don't see how to chop a corner off a square and obtain a regular polygon with eight sides,?
– JKEG
May 10, 2018 at 18:57
• @JKEG s/he meant all corners
– SK19
May 10, 2018 at 18:59
• Please use MathJax. May 10, 2018 at 18:59
• @SK19 So this question is "what's the area of an octagon"?
– JKEG
May 10, 2018 at 19:03
• @JKEG No, it's "I'm trying to find the area of an octagon, how can I proceed with my method?" which I convey as different, as you can see in my answer.
– SK19
May 10, 2018 at 19:09

## 3 Answers

If the side of the cut triangle is $x$, then in order to have a regular polygon, we need to have $$1-2x = \sqrt 2 x$$

solving for $x$ we get $$x= 1- \frac {\sqrt 2 }{2}$$

The total area of the cut is $2x^2$ and the area of polygon is $$A=1-2x^2 = 2 (\sqrt 2 -1) \approx 0.8284$$

An image makes everything simpler to understand:

Because the square side length is 1, $$b + a + b = 1$$ Solving for $b$ we get $$b = \frac{1 - a}{2} \label{NA1}\tag{1}$$ Squaring this (noting that we are limited to positive $a$ and $b$, i.e. that $a \gt 0$ and $b \gt 0$) we get $$b^2 = \frac{1 - 2 a + a^2}{4} \label{NA2}\tag{2}$$

In the corners, we need $b$ such that the hypotenuse is $a$. Pythagorean theorem says $$b^2 + b^2 = a^2$$ solving for $b$ we get $$b^2 = \frac{a^2}{2} \label{NA3}\tag{3}$$

Combining $\eqref{NA2}$ and $\eqref{NA3}$ we get $$b^2 = \frac{a^2 - 2 a + 1}{4} = \frac{a^2}{2}$$ i.e. $$2 a^2 = a^2 - 2 a + 1$$ which simplifies to $$a^2 + 2 a - 1 = 0$$ and noting that the first two terms are from squaring $a$, to $$a^2 + 2 a + 1 - 2 = (a + 1)^2 - 2 = 0$$ Moving the last term ($-2$) to the other side, we get $$(a + 1)^2 = 2$$ so taking a square root on both sides, and remembering that we need $a \gt 0$, we get $$a = \sqrt{2} - 1$$ Substituting this to $\eqref{NA1}$ we can solve $b$, $$b = \frac{1 - a}{2} = \frac{1 - \sqrt{2} + 1}{2} = \frac{2 - \sqrt{2}}{2} = 1 - \sqrt{\frac{1}{2}}$$ Note also that $$b^2 = \left(1 - \sqrt{\frac{1}{2}}\right)^2 = 1 - 2 \sqrt{\frac{1}{2}} + \frac{1}{2} = \frac{3}{2} - \sqrt{2}$$

The corner triangles total area is the same as the area of two $b$-sided squares. So, the area of the octagon is the area of the unit square (1) minus the area of two $b$-sided squares: $$A_{octagon} = 1 - 2 b^2 = 1 - 2 (\frac{3}{2} - \sqrt{2}) = 1 - 3 + 2 \sqrt{2} = 2 \sqrt{2} - 2 \approx 0.828$$

Let's have a look at the corners we chopped off. It is a right triangle, because it was the corner of a square. It is an equilateral triangle because of symmetry of the resulting polygon. Its hypotenuse is the base of the triangle you are looking for.

Lets call the length of the hypothenuse $b$ and the length of each of the other two sides $a$, then by Pythagoras

$$b^2 = 2a^2$$

On the other hand, the side of your square is such a base of a triangle into which you divided the polygon + 2 times the length of the shorter side of the triangle you chopped of (that is, $a$), so if $c$ would be the length of each site of the original square, we get

$$c=b+2a$$

As I understand you, $c$ is given (I think you mentioned $c=1$ in your case) so you can effectively solve for $b$. (note that $2a^2 = 2(a^2) \neq (2a)^2$ in general)

Answering your question from your text aside, the figure you are trying to find the area for is an octagon, so it's area would be $2(1+\sqrt{2})b^2$ or, equally, be $2(\sqrt{2}-1)c^2$ (since $c$ is double the apothem, see the wikipedia link).