Yes, squares and rectangles are easy things for which to calculate area. Just use base times height. And triangles are almost the same. Just multiply base times height times one-half. But what about other shapes?
First, a note on vocabulary and notation. In this entry, I will be showing how to get the formula for the area of any regular convex polygon. A polygon is a shape with at least three sides. A regular polygon has all sides of the same length and all angles of the same measure. A convex polygon has each angle measuring less than 180 degrees.
The symbol s always refers to the length of a side.
The symbol s/2 always refers to half of the length of a side.
The symbol p always refers to the perimeter of a figure.
The symbol n always refers to the number of sides.
The symbol A always refers to an area.
The symbol a always refers to the apothem. The apothem is a line drawn from the center of the figure to a side. The apothem cuts the side into equal halves (bisects) at two 90 degree angles (perpendicular).
Finally, a note on the writing. As I said, I'm talking about regular polygons, so all sides within a polygon are the same length (even if they do not appear to be so). The drawings may look a little rough, but they are legible and should drive the point across.
Let's begin by looking at the simplest polygon: the triangle.
This triangle has three equal sides, all of length s. Each dotted line represents an apothem. The details are similar for the square and hexagon below, except the square has four sides, and the hexagon has six sides.
In general, we refer to a polygon that has n sides as an n-gon.
From the images above, we can notice a pattern in the relationship between the number of sides in a polygon and the number of little triangles created with lines drawn through each angle to the center and lines drawn from the center to the sides. Please note: in general, it is not safe to assume a trend from a few examples. For the purposes of this exercise, we just need to convince ourselves that the trend below is true (an n-gon has 2n triangles).
In the case of the square, we can calculate the area of the shaded triangle below using the formula for the area of a triangle, where the base is s/a and the height is the apothem.
The same is true for the hexagon.
In fact, the area of a little triangle of an n-gon is also A = sa/4. Again, we cannot, in general, make such assumptions from a few examples. If we can convince ourselves this is true, it will suffice for our purposes.
Now to find the area of the entire square, we just need to multiply the area of one little triangle by the number of little triangles. Since these are regular convex polygons, each little triangle of the shape contains the same area. As we said in the case of the square, there are 8 little triangles. So the total area is 2sa.
In the case of the hexagon, there are 12 little triangles, so we get a total area of 3sa.
For an n-gon, as we said earlier, we have 2n little triangles. This gives us a total area of (2n)(sa/4). This simplifies to nsa/2.
Perimeter is simply the sum of the lengths of the sides. For the square, there are four sides of length s, so the perimeter is 4s. In other words, one side is one-fourth of the perimeter.
For the hexagon, there are six sides of length s, so the perimeter is 6s. In other words, one side is one-sixth of the perimeter.
For an n-gon, there are n sides of length s, so the perimeter is ns. In other words, one side is one-nth (1/n) of the perimeter. Or, s = p/n.
We can combine our equations for area with the equation relating side length and perimeter. Since, in the case of the square, s = p/4, we can plug p/4 into the Area = 2sa equation.
The process is similar for the hexagon.
We repeat the process one final time for the n-gon.
In all three cases, we came up with the formula Area = (1/2)(a)(p). While this was not a rigorous proof, it was a way to see where the formula comes from. And while not very interesting on its own merits, it will be quite useful in answering tomorrow's question. I just wanted to devote an entry to this problem because it is worth seeing.
Anyway, I hope you got something out of this. If not, well, there's always next time.
If you have any questions, or if anything isn't clear, just let me know in the comments.
I first saw this formula in my freshman year of high school in Mrs. J. Plunkett's Geometry I class. I didn't see any sort of derivation until I worked one out myself the summer following my sophomore year of high school.
The images here were produced using Noteshelf for iPad.
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