1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't see how to chop a corner off a square and obtain a regular polygon with eight sides,? $\endgroup$ – JKEG May 10 '18 at 18:57
  • $\begingroup$ @JKEG s/he meant all corners $\endgroup$ – SK19 May 10 '18 at 18:59
  • 1
    $\begingroup$ Please use MathJax. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 10 '18 at 18:59
  • $\begingroup$ @SK19 So this question is "what's the area of an octagon"? $\endgroup$ – JKEG May 10 '18 at 19:03
  • $\begingroup$ @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. $\endgroup$ – SK19 May 10 '18 at 19:09
2
$\begingroup$

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 $$A=1-2x^2 = 2 (\sqrt 2 -1) \approx 0.8284$$

$\endgroup$
2
$\begingroup$

An image makes everything simpler to understand: Unit square and octagon

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: $$A_{octagon} = 1 - 2 b^2 = 1 - 2 (\frac{3}{2} - \sqrt{2}) = 1 - 3 + 2 \sqrt{2} = 2 \sqrt{2} - 2 \approx 0.828$$

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.