Tuesday, March 4, 2014

I/D# 2: Unit O Concepts 7-8: How can we derive our patterns for our speical right triangles?


Inquiry Activity Summary:

In class we were given a square with side lengths of 1 and an equilateral triangle, also with side lengths of 1. We were given directions to derive the patterns for the 45-45-90 and 30-60-90 triangles from the square and the equilateral triangle using only what we knew of the two shapes (ignoring any pre-knowledge of the 45-45-90 and 30-60-90 patterns). The following is how I accomplished this heroic task:

1). Deriving the 30-60-90 pattern with an equilateral triangle:

First things first, you want to draw an equilateral triangle. We know the sides are all equal to 1 and we know that the three angles are all equal to 60 degrees in this equilateral.
Now in order to begin our tedious construction of a 30-60-90 triangle, we will slice the above triangle equally in half right down its altitude (top middle). Several things will change; the first and most obvious being that we will have 2 triangles. Second, the top 60 degree angle will be cut into two 30 degree angles (one for each triangle). And lastly, the bottom side of the original triangle will also split in half, resulting in two horizontal side lengths of 1/2 (again, one for each triangle). Also important is that the intersection cuts the triangle perpendicularly, meaning that you will have two 90 degree angles on each of the two new triangles. [Note: If you haven't noticed by now, we will be deriving not one 30-60-90 triangle, but two.] Take notice that we already have the 30-60-90 degree relationship established by this step.
Next on the agenda would be to find our missing side: the vertical value which is known as "y". We will use only one of the two triangles (doesn't matter which) and the Pythagorean theorem to find that missing side. We do the math and we find that the missing side (y) is equal to (radical 3)/2. This is demonstrated below. We now have derived every part of the 30-60-90 triangle but we are still not done just about yet.
The side values we have on the triangle right now, although they are correct, are not what we are used to seeing in the 30-60-90 pattern. We must do away with the fractions and we do this by multiplying ALL the sides by 2. The results are shown below:
Now all that's left is to place in "n" by multiplying it to ALL sides. We do this because since we are deriving a pattern, we don't want just a set of numbers, we want a set of constants that make up the rule of the formula. The below picture is the 30-60-90 triangle that we have fully derived from the equilateral:

2). Deriving the 45-45-90 pattern from a square

Okay. First I drew the square. Its sides will be one and like all squares this one will have 4 right angles.
 Next we cut the square equally down its diagonal. The results will be two right triangles, and as shown, two of the right angles will be cut in half into two 45 degree angles. We can already see the 45-45-90 degrees in out triangle, and we have two of the sides; we just need that third missing side again. But also take note that the two sides, like the two angles, are equal; this is because the sides correspond with the angle opposite to it and since they are equal angles, there will be equal sides.

Now we use the Pythagorean theorem to find that missing side. Again this is demonstrated below. We shall find that the missing side (this time "r") is equal to radical 2.
Now we multiply by "n" to get our constant, which represents the overall relationship and pattern that this special triangle follows.
And now we can separate the triangle and see that we have fully derived the 45-45-90 triangle pattern from the square.

Inquiry Activity Reflection:

1). Something I never noticed before about special right triangles is... that they come from a square and an equilateral triangle, and that the patterns they follow are not just numbers that happen to work every time, but they actually came from what we did in this activity.

2). Being able to derive these patterns myself aids in my learning because... learning something isn't just memorizing something with no meaning; this helps me understand what special right triangles really are.



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