area of a regular polygon

For example, a regular polygon is a square - it is a polygon that has all sides the same length and all angles the same size. For example, this is what a regular octagon looks like:

How would we calculate the area of such a regular octagon if we know the length of its side? For example, suppose a side has length a = 10. We can help by dividing the octagon into eight congruent triangles:

This has simplified our work - now we just need to calculate the area of one of the triangles and multiply this content by eight to get the area of the whole octagon. What do we know about these triangles? They are isosceles triangles - the lengths of the two sides that point away from the centre S are always the same length, and the length of the side of the octagon is different (only the hexagon can be divided into triangles that have all sides the same length). Thus, the length of the side of FS is the same as the length of the side of ES.

What angle do they make at the vertex S? That is, what is the angle of $\angle FSE$, for example ? We can see that all triangles are the same and therefore all angles at the vertex S must be the same size. At the same time, we know that if we add the magnitudes of all these angles, we get 360 degrees - one full revolution. So the size of the angle $\angle FSE$ must be equal to 360 divided by the number of angles, i.e., divided by the number of triangles.

$$\Large \angle FSE = \frac{360^\circ}{8} = 45^\circ$$

Since the sum of all interior angles in a triangle is equal to $180^\circ$, and since we have an isosceles triangle, it must be true that the other two angles have the same magnitude $67,5^\circ$:

Now how do we calculate the area of an isosceles triangle? We know the length of the side FE from the assignment, that is |FE| = 10. We now draw the height of the triangle that passes through the point S:

This line SP_s has divided our triangle into two smaller triangles, but they are the same again - they are just mirrored. Let's rearrange the triangles a bit. Let's take the triangle FSP_s and move it so that the pattern forms a rectangle:

This rectangle SF₁EP_s has the same content as our triangle FSE. We calculate the area of the rectangle as the product of the lengths of its two sides:

$$\Large S_\square = |v_s|\cdot |P_sE|$$

We know the length of the line segment P_sE, it is half the length of the side FE and it has the length (according to the assignment - it is the side of the octagon) |FE| = 10. The line segment P_sE thus has the length |P_sE| = 5. It is worse with the height v_s. Since the triangle P_sSE is a right triangle, we can use the goniometric function. The angle $\angle P_sSE$ is half the size of the original angle $\angle PSE$, so

$$\large \angle P_sSE = \frac{\angle PSE}{2}=\frac{45^\circ}{2}=22{,}5^\circ$$

We now use the fact that the cotangent of an angle is equal to the ratio of the length of the adjacent branch to the opposite branch. For clarity, we will denote the angle $\angle P_sSE$ by the Greek letter α:

$$\Large \mbox{cotan}\ \alpha = \frac{|v_s|}{|P_sE|}$$

We need to find |v_s| from this equation, so we multiply the whole equation by |P_sE|, to get:

$$\Large \mbox{cotan}\ \alpha\cdot|P_sE|=|v_s|$$

Now we know what height equals. Let's add up the values:

$$\begin{eqnarray} |v_s|&=& \mbox{cotan}\ \alpha\cdot|P_sE|\\ &\approx&2{,}414\cdot5\\ &\approx&12 \end{eqnarray}$$

Height has a length of approximately 12. The area of a rectangle, and therefore a triangle, is equal to

$$\Large S_\triangle=|v_s|\cdot|P_sE|$$

After offsetting, we have:

$$\Large S_\triangle=12\cdot5=60$$

The area of a triangle is equal to approximately 60. Since we have a total of eight triangles, the area of a regular octagon with side length a = 10 is equal to approximately

$$\Large S=8\cdot S_\triangle = 480$$

How do we get the formula from this? For a general n-rectangle with side length a, the following would be true

$$\large S = n\cdot S_\triangle$$

And that leaves us with the general formula for $S_\triangle$:

$$\large S_\triangle = \frac{a}{2} \cdot \mbox{cotan}\ \alpha\cdot\frac{a}{2}$$

The angle α is half the magnitude of the angle that, thus

$$\large \alpha = \frac{360^\circ}{n\cdot2}=\frac{180^\circ}{n}$$

We add back:

$$\large S_\triangle = \frac{a}{2} \cdot \mbox{cotan}\ \frac{180^\circ}{n}\cdot\frac{a}{2}$$

And we just need to smooth it out a bit:

$$\large S_\triangle =\frac14\cdot a^2 \cdot \mbox{cotan}\ \frac{180^\circ}{n}$$

In the last step, we just multiply this value by the number of sides in the n-angle and we have the resulting formula:

$$\large S =\frac14\cdot a^2\cdot n \cdot \mbox{cotan}\ \frac{180^\circ}{n}$$

Calculate the area of a regular n-gon.