8.4
Regular Polygons
While there are no general methods for computing areas of most polygons, the situation is quite different when it comes to regular polygons. In a regular polygon, all the angles are congruent and all the sides are congruent. If we take $n$ uniformly spaced points around a circle (where $n\geq 3$) and join them with segments, we know that those segments are all congruent (because congruent arcs have congruent chords) and the angles are all congruent (because each is an inscribed angle intercepting $n-2$ of the congruent arcs). Hence we have a regular polygon.
Definitions: | Given a regular polygon, the center of the polygon is the center
of the circumscribing circle; the radius
of the polygon is the radius of the circumscribing circle; the apothem of the polygon is the
distance from the center to a side. |
Note that
when we join the center of a regular $n$-gon to each of the vertices,
we obtain $n$ congruent isosceles triangles. Pictured to the
right is a regular pentagon; each central angle is
$\left(\frac{360}{5}\right)^\circ$, and each of the five triangles
formed has radii of the circle for two of the sides. To find the
area of the polygon, we need to find the area of one of the triangles and multiply by
the number of sides. Another way to get at this idea is to use
the following theorem. |
- Theorem 8.6:
The area of a regular polygon is one-half the product of the apothem
and the perimeter.
Given:
$ABCD\ldots$ is a regular polygon with $n$ sides. The apothem is
$h$; the perimeter is $p$. Prove: $a(ABCD\ldots)=\frac{1}{2}h\cdot p$ Proof: Draw the radii of the polygon, creating $n$ congruent triangles. If the length of a side of the polygon is $s$, then the area of each triangle is $\frac{1}{2}h\cdot s$. Since the polygon's area is the sum of $n$ such triangles, the area of the polygon is \begin{align} a(ABCD\ldots)&=n\left(\frac{1}{2}h\cdot s\right)\\ &=\frac{1}{2}h\cdot (ns)\\ &=\frac{1}{2}h\cdot p \end{align} |
Example:
|
A regular octagon has a side of
2 and an apothem of $1+\sqrt{2}$. What is the area of the octagon? |
|
Solution:
|
The perimeter is 16. Therefore the area is given by $\text{area }=\frac{1}{2}(1+\sqrt{2})(16)=8+8\sqrt{2}$. |
It is important, when dealing with a regular polygon, to check what each of the central angles is. For example, a regular polygon with 18 sides (known as an 18-gon) has central angles of $20^\circ$. In problem 22 of Exercises 8.2, we proved that the area of a triangle with sides $a$ and $b$ including $\angle C$ has an area given by $\frac{1}{2}ab\cdot \sin \angle C$. If an 18-gon is inscribed in a circle of radius 5, it can be thought of being made up of 18 separate triangles, each with an area of $\frac{1}{2}\cdot 5\cdot 5\cdot \sin 20^\circ$. Then the area of the 18-gon is $18\cdot (\frac{1}{2}\cdot 5\cdot 5\cdot \sin 20^\circ)=76.95$.
There is another approach that can lead to the exact calculation of the area of a regular polygon in certain cases. If the polygon has central angles of 45° (an octagon) or 30° (a dodecagon), the apothem is hard to calculate exactly, but putting in a different altitude in each of the triangles that make up the polygon takes advantage of the nice central angle.
Example:
Solution: |
Determine the area of a regular
dodecagon of radius 6. The central angle for each of the 12 central triangles is $30^\circ$. In $\triangle ABO$, put in the altitude from $B$, as shown (not the apothem). Then $\triangle BOK$ is a 30-60-90 triangle, and $BK=3$. Hence $a(\triangle AOB)=\frac{1}{2}(6\cdot 3)=9$. That gives the area of the dodecagon as $12\cdot 9=108$. |
This approach works nicely when the radius of the octagon or the dodecagon is known. When dealing with a regular octagon with a known side, two other schemes are better. One is to break the octagon into triangles, rectangles, and a square, as pictured to the right. If the side of the regular octagon is 4, there are four right triangles with legs of $2\sqrt{2}$, four rectangles with sides of 4 and $2\sqrt{2}$, and a central square with a side of 4. The total area is then $32+32\sqrt{2}$. |
Even easier is to picture a square around the octagon. The side of the square is $4+4\sqrt{2}$. To get the area of the octagon, we find the area of the large square and subtract the four outside triangles. This gives $(4+4\sqrt{2})^2-4\cdot 4=32+32\sqrt{2}$. |