# Prime Eval, Part 4

In part 3, I jumped from one to two dimensions, and invoked an x-y axis system, thereby kind of getting ahead of myself.

But, when we switch to geometry, we need to deal with shapes, and the first real place we can do that is when we go to a 2D plane. A line segment has no width, so the shape it forms will have no area. Same holds true if we have a shape formed by 2 line segments. It’s when we hit three segments, or a shape with 3 sides, that we can actually say we have a shape and that the shape has an area. I’m pretty sure that the ancient Greeks would like to claim to have made all the discoveries there are about triangles, such as the angles adding up to 180 degrees and all that, and maybe they did (according to the wiki entry, they didn’t, and that’s good enough for me). Anyway, I wasn’t there at the time, so I remain skeptical. But one thing that they (allegedly) observed is that triangles that have one angle of 90 degrees were a lot easier to work with than the other ones are.

Say you have a random triangle and you want to know what its area is. You could try doing all kinds of tricks if you like. But, finding the two lengths created by drawing the perpendicular line on the longest leg so that it intersects the opposite corner will give you two smaller right angle triangles. How do we define “perpendicular”? By saying that the angles created on each side of the line are equal. Since the line of the longest leg is straight, it represents a 180-degree angle at the start. The only way the two angles on each side of the perpendicular line can be equal (in flat 2-D space) is if they’re both 90 degrees. Put four of these angles at the corners of a test object and we get a four-side thingie we call a rectangle. From this, we can see that our random triangle can be divided into two smaller right triangles, which are then each half the areas of the two rectangles formed by multiplying the lengths of each of their sides together.

So, for a random triangle, we can find two smaller right-angle triangles within it. Take the lengths of the two shorter sides for each new triangle and multiply them together and divide by two to get the areas of each right-angle triangle. Then add the two smaller triangle areas together to get the final answer for the area of the original triangle.

Add a side. Get some kind of random 4-sided thingie. What’s the area? Well, draw a line from one corner to the opposite one, then break the two resulting triangles into 4 right-angle triangles. Multiply the two sides of each, add them together, and divide by 2. Doesn’t matter what the initial shape is, if we cut it up so that we have something with 3 sides, and one angle is 90 degrees, the math becomes really simple. And, of course, what I’m leading to is the special case represented by a rectangle, where all four corners are right angles and the 2 pairs of opposing sides are of equal length. Which then ties into the perpendicular lines used to draw the x-y graph. It’s all very arbitrary, but if you ask the question “are mathematicians lazy”, and you get the answer “yes” for at least one of them, then doing the math in Cartesian space with right-angle triangles and rectangles is going to save you a lot of otherwise needless work.

One thing that I hadn’t seen when I was in my high school math class was the following representation of the Pythagorean Theorem. We all know (I hope) that Pythagy (as he was called by his friends, although he kept saying he hated that name) was the first person on permanent record as saying that for a right-angle triangle, the square of the length of the longer leg is the sum of the squares of the two shorter ones. But, what exactly does that mean?

Take a triangle. If we take a second triangle of the same dimensions and flip it around, we’re going to make a rectangle. The area of the rectangle is the product of its length and width. Therefore, the area of the triangle will be one half of the rectangle it forms. We want to connect that area somehow to the length of the longer leg (the hypotenuse). What Pythagy realized was that the two other legs could create squares themselves, and that squares are just a special case of the rectangle.

Digression: Is there a relationship between a rectangle and a square? Well, if we draw out the rectangle on graph paper, and then square both sides, we get something that looks like this: 4 things that when next to each other like this form a bigger square. Say the original rectangle was 4×3. 4×4 = 16. 3×3 = 9. and the rectangle itself shows up twice, as 3x4x2 = 24, for a total area of 49. While the big box now has a side of 7, or the square root of 49… (a+b)*(a+b) = a^2 + 2ab + b^2.

Umm… Maybe. I’d swear I’ve seen that formula before…

End of digression. Anyway, this is what I’ve seen: The areas of the squares created by the two shorter legs added together equals the area of the square created by the hypotenuse. I mean, yeah, it’s just a visual representation of c^2 = a^2 + b^2. But I don’t remember ever seeing that when I was in school. It makes a lot of sense now. I just don’t understand why I can’t remember having seen it when I was in school… On the other hand, Pythagy’s proof was a bit more straightforward to what I’m used to.

(Image from the wiki article)

If you lay out a triangle with sides A and B as shown in the left image, you get a square with a length of C, and an area of C^2. Rearranging the triangles to make the image on the right is functionally equivalent, but it gives you two smaller squares of areas A^2 and B^2. Since the space in white has the same area in both images, it’s apparent that the areas of A^2 plus B^2 are equal to the area C^2. The total space is (a+b)^2 again, and the only difference between what I’d drawn for my rectangle and the Pythagorean Theorem is that we’re subtracting out the 2ab part. That is, the total space in the left image has an area of c^2 + 4*(a*b/2) = c^2 + 2ab. While the total space on the right is a^2 + b^2 + 2*ab. (Just add up each of the squares and triangles). Since the total areas of both images are equal, c^2 + 2*ab = a^2 + b^2 + 2*ab. Or, c^2 = a^2 + b^2.

I’ve been flogging a dead horse here, but I just want to get the basics out of the way before proceeding any further. (Plus, it wasn’t my horse.)

Previous Post

• ## Follow TSOJ in email

Join 101 other followers

• ## Blog Stats

• 63,851 hits