Apothems and Area

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What is an Apothem?

The image to the right shows the shortest distance from the center to the midpoint of one side in various regular polygons.


Contents

Basic Description

An apothem extends from the center of a regular polygon to the midpoint of one of its sides. If you know the lengths of the apothem and one side of a regular polygon, you can easily find its area. If the regular polygon (see the hexagon in Figure 1) is divided into triangles, the triangles can be unrolled to form half of a rectangle.


Figure 1
Figure 1


Figure 2
Figure 2

Let’s start with a simple example. We will use the apothem of a hexagon to find its area as shown in Figure 2.

  1. Start with two hexagons.
  2. Assume that each side is x units long. The perimeter of each hexagon is 6x units.
  3. Divide each hexagon into six equilateral triangles (illustrated below). Each equilateral triangle has sides of length x and a height equal to the apothem. Let’s call this a.
  4. "Unwrap” each hexagon to get two rows of congruent equilateral triangles. These two rows fit together nicely in a rectangle. The base of the rectangle is equal to 6x, so the area of the rectangle is equal to 6xa. But what is a? To find out the area of the rectangle and therefore the hexagon, we need the rectangle's height, which is equal to the apothem. Since the apothem cuts each equilateral triangle into two right triangles we can use trigonometry to solve for a.


Figure 3
Figure 3
cos30^\circ = \frac{a}{x}

 a = xcos30^\circ

Thus the area of the rectangle is  6x \cdot xcos30^\circ
The area of the hexagon is equal to half the area of the rectangle so the area of the hexagon is equal to
3x^2cos30^\circ


A More Mathematical Explanation

Note: understanding of this explanation requires: *Geometry, Basic Algebra

General Formula for Finding Area

Consider a regular polygon with n sides that are each x [...]

General Formula for Finding Area

Consider a regular polygon with n sides that are each x units long. Let’s find a formula for the apothem (a) that will help us to find the area. Since the figure is a regular polygon, the measure of one of its angles is

\frac{(n-2)180^\circ}{n}.

How do we know this? Let's derive a general formula by looking at a regular hexagon. Draw a line from one vertex of the hexagon to the other 4, non-adjacent vertices.

Figure 4
Figure 4
We see that this forms 4 triangles. Each triangle's angles must total 180 degrees. Therefore, the total degrees of the internal angles of the triangles must be
4 \cdot 180 = 720^\circ .

And since the hexagon has 6 equal angles, each angle must be equal to

 \frac{720}{6} = 120^\circ .

We can generalize this method over all n-polygons divided into n-2 triangles to say that the measure of each angle in an n-sided polygon is equal to

\frac{(n-2)180^\circ}{n}.
Figure 5
Figure 5

Refer to Figure 5. We've connected center A to vertices B and C to create a triangle. We then dropped an altitude, creating line segment AD whose length is the apothem, a. So we want to know as much about triangle ABD as possible. The first thing to get would be the measure of angle ABC in terms of x, n, or a. We can do this because we already have angle MBC's measurement from the formula above:

\frac{(n-2)180}{n}.

ABC is half of MBC

We know this because the triangles we've divided the regular polygon into are congruent isosceles triangles. The base angles of an isosceles triangle are congruent to each other, and all the triangles that make up the polygon are congruent. Therefore, the base angles of all the isosceles triangles in our regular polygon are congruent. Because angles MBA and ABC are base angles of isosceles triangles in our polygon, they are congruent. As they are congruent and add up to angle MBC, the measure of each is half the measure of angle MBC.

so we just multiply that expression by 1/2 to get

\frac{(n-2)180}{2n}

So now we have the measure of angle ABC, expressed in terms of n. What else do we know about triangle ABD? Well, as line segment BD is half of BC, and BC's length is x, BD's length is .5x.

We also know that angle ADB is a right angle, thus making ABD a right triangle. This is important because right triangles have special properties like the tangent ratio. We can (and will!) use this ratio with our other two pieces of information. The tangent of angle ABD is the ratio of the opposite side over the adjacent side--in this case, a over BD. We have the measure of BD already, so we can set up an equation and solve for a!


Figure 6
Figure 6
Look to Figure 6. By using the equations in the red box (which is what we know about the triangle) we can substitute in the values we know in terms of x and n for the parts of the triangle they represent. We now have a formula for the apothem: a=\frac{1}{2}x\cdot\tan\left(\frac{(n-2)180}{2n}\right)

How can we check this formula? Well, with a polygon whose apothem we can check with other ways. A regular hexagon can be formed from six triangles that are not only isosceles but also equilateral. This will be important later. For now, let's try it out...

If we plug the values for a hexagon with side length 1 (for simplicity) into our formula, we get the following:

a=\frac{1}{2}(1)\cdot\tan\left(\frac{(6-2)180}{2(6)}\right) which simplifies into roughly .866.

Let's return to the image above. Because we're using a hexagon, triangle ABC is equilateral, so triangle ABD is a 30-60-90 triangle. In a 30-60-90 triangle, the shorter leg (BD) is s, the hypotenuse (AB) is 2s, and the longer leg (a) is \sqrt{3} times s. Our hexagon has side length 1, so .5x is just .5. If we multiply this by \sqrt{3}, we get around .866. Our formula works!




Using the Apothem to Solve the Wire Problem

A common optimization problem encountered in calculus classes involves cutting a wire in half and morphing each half into a certain shape. The wire problem asks students to cut the wire so that the sum of the enclosed areas is maximized or minimized. Students are often asked to model the following scenarios:

  1. a circle and a square
  2. a circle and an equilateral triangle
  3. an equilateral triangle and a square

We will go through a minimization example of the wire problem below. Say we have a 2 foot piece of wire that will be bent into a square and an equilateral triangle, and we want to minimize the sum of the enclosed area. Where should we cut the wire?

Figure 7
Figure 7

We will let the wire for the square be x feet and the wire for the triangle be 2-x feet. Each side of the square will be x/4 and each side of the triangle will be 1/3(2-x). The height of the triangle will be √3/6(2-x) So the total area of both objects
A(x)= (\frac{x}{4})^2 + \frac{1}{2}(\frac{1}{3}(2-x))(\frac{\sqrt{3}}{6}(2-x))= \frac{x^2}{16} + \frac{\sqrt{3}}{36}(2-x)^2
Take the first derivative of the area
 A'(x)= \frac{x}{8} + \frac{\sqrt{3}}{6}(2)(2-x)(-1)=\frac{x}{8}-\frac{\sqrt{3}}{9}+\frac{\sqrt{3}}{18}x
Set this equal to zero:
0 =\frac{x}{8}-\frac{\sqrt{3}}{9}+\frac{\sqrt{3}}{18}x
Solve to get:
x = \frac{8\sqrt{3}}{9+4\sqrt{3}} = 0.8699
The above is a critical point. This means that the sum of the enclosed area of the two shapes must be either maximized or minimized when x=0.8699. It appears that the sum of the enclosed area is minimized at this point, so
Wire for square: 0.8699ft
Wire for triangle: 2-0.8699 = 1.1301 ft


If you are unfamiliar with the wire problem, example 4 on this webpage provides a more detailed explanation of the above problem.

Pat Cade and Russell A. Gordon of Whitman College have demonstrated a unique way to solve the wire problem in their paper, An Apothem Apparently Appears.

They have found that in each minimization solution one shape is inscribed in the other.

  1. the circle will be inscribed in the square
  2. the circle will be inscribed in the equilateral triangle
  3. the equilateral triangle and square can be inscribed in the same circle

While scenario 3 is slightly different, in all three scenarios the two shapes have the same apothem! This enables us to take some shortcuts in solving the wire problem.

Figure 8
Figure 8

How did they do this? First, the perimeter P and area A of a regular n-gon with apothem r are given by
P=2\alpha_nr
A=\alpha_nr^2 where
\alpha_n=ntan(\frac{\pi}{n})
Say we have two shapes, one p-gon and one q-gon. It is possible that p=q. The sum of the perimeters of the two shapes will be equal to L and the sum of the areas will be equal to Q.
L=2\alpha_px+2\alpha_qy
If we take the derivative with respect to x we get the following,
0=2\alpha_p+2\alpha_q\frac{dy}{dx}
Simplifying the above equation gives us:
\frac{dy}{dx}=\frac{-\alpha_p}{\alpha_q}
Now let's look at the sum of the areas:
Q=\alpha_px^2+\alpha_qy^2 subject to x \geq 0, y \geq 0, L=2\alpha_px+2\alpha_qy
We want to take the derivative of the above equation for Q with respect to x:
0=2\alpha_px+2\alpha_q\frac{dy}{dx}= 2\alpha_px-2\alpha_qy\frac{\alpha_p}{\alpha_q}=2\alpha_p(x-y)
Simplifying this gives us:
0=2\alpha_p(x-y)
Once again, we have a critical point. This time it is at x=y or when the shapes have the same apothem.


For more information and to read the paper click here!
If you cannot access the above link, try here for the first 3 pages.


Eliminating Calculus

Knowing that the two apothems must be equal in order to minimize area eliminates calculus from solving basic scenarios 1, 2, and 3. Say, for example we are solving scenario 1. The apothem of the circle will be equal to the apothem of the square. The apothem of the circle is equal to the radius; let’s call this r. The apothem of the square will also be equal to r so each side of the square will be equal to 2r.

Figure 9
Figure 9
Circumference of Circle =  2\pi r

Perimeter of Square =  8r
Since we know the total perimeter of both shapes combined must equal the original wire length (let’s call this L) we know that
L = (8r) + (2\pi r) . This can be rewritten as
L = (8+2\pi)r.
All we need to know is the length of the wire, L, and we can easily solve for r.
r = \frac{L}{8+2\pi}

Once we know r we can use this value to calculate the circumference of the circle, and use that value to find the appropriate cutting point on the wire.
If the circumference = 2 \pi r then the wire must be cut at L-2\pi r

Figure 10
Figure 10


What if the wire problem involves a triangle? We can still eliminate calculus from the problem, but it is slightly trickier. Note that the apothem of the triangle (labeled in Figure 11 as r1) is now equal to 1/3 the height.

Figure 11
Figure 11
The centroid of an equilateral triangle is the intersection of the medians of the triangle . The median of an equilateral triangle is congruent to the apothem. . [1]

Because the height divides the original triangle into two right triangles, we can use trigonometry to find the value of x, and then easily find the perimeter of the triangle.

Figure 12
Figure 12

tan(30) = \frac{x}{3 \cdot r_1}

x = tan(30) \cdot 3 \cdot r_1

Perimeter of triangle = 6 \cdot x

Perimeter of triangle = 18 \cdot r_1 \cdot tan(30)

Next we need to set the sum of the perimeters equal to L1


Circumference of the circle = 2\pi r_1

Perimeter of triangle = 18 r_1 tan(30)

 L_1 = (2\pi r_1) + [18 r_1 tan(30)] factor out the r to get

 L_1 = r [2\pi + 18tan(30)] which can be written as

 r = \frac{L_1}{2\pi + 18tan(30)}

Figure 13
Figure 13

All we need to know is the length of the wire (L1) and we can solve for r and find the optimum place to cut the wire (just like the problem above).

Fun fact: the apothem of the Pentagon is about 633.8 feet!
Fun fact: the apothem of the Pentagon is about 633.8 feet!




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References

National Security Agency's K-12 Academia Program
  1. [ http://en.wikipedia.org/wiki/Centroid "Centroid"], Retrieved on 16 July 2012.





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