I/D1: Unit O - Derive the Special Right Triangles

Inquiry Activity Summary
1. To derive the 30-60-90 triangle from a equilateral triangle with all side lengths of 1, we first have to cut the triangle in half down the middle, perpendicular to the base side. The angles that will have been formed in the new triangle (either half of the divide) are 30, 60, and 90 degrees. The 90 degrees is formed from the perpendicular cut and the base of the triangle, the 60 degrees is the untouched angle of the new triangle, while the 30 degrees angle has been the angle cut in half where the divide had begun at (half of 60 degrees is 30 degrees).
For the side lengths, the hypotenuse is still one, since that was the original length of a side of the equilateral triangle. The 30 degree side is 1/2, since the triangle had been cut in half from the base and we are deriving half of the original side length of 1. Finally, the 60 degree side can be derived using the Pythagorean theorem. We must solve for 'b' once we plug in the numbers for 'a' and 'c' (1/2 and 1, respectively, corresponding to their angles). When we simplify and get the 'b' variable on its own, b^2 will be equal to 3/4. As a result, the value of b would be radical 3 over 2. Finally, we must multiply all the side lengths by 2 to simplify them, since we do not want fractions. The new side lengths would be, in ascending order: 1, radical 3, and 2. The last step is to substitute the values with the 'n' variable where n=1. This shows that these values can be applied to any multiples of themselves. Now, the new side lengths would be, in ascending order, 'n' to the 30 degree side, n radical 3 to the 60 degree side, and 2n to the 90 degree side.

2. To derive the 45-45-90 triangle from a square with all side lengths of 1, we first have to cut the square in half diagonally, from one corner to another. The angles that will have been formed in the new triangle (either half of the divide) are two 45 degree angles. This is because the two 90 degree angles that have been cut in half are both now 45 degrees, since half of 90 degrees is 45 degrees. The third and last angle is 90 degrees, which is the corner of the former square that had been untouched. These angles fulfill the 180 degree rule for all triangles.
For the side lengths, the two legs of the 45-45-90 triangle are both one, since they were from the sides of the square. If we plug this into the Pythagorean theorem, we see that c, or the hypotenuse, is radical 2. This is because c^2=1+1=2, so we take the radical of it. So, our three side lengths would be, in ascending order, 1, 1, and radical 2. Then, we substitute the values with the variable 'n', where n=1. This shows that these values can be applied to any multiples of themselves. Now the new side lengths would be 'n', 'n', and n radical 2.

Inquiry Activity Reflection
Something I never noticed before about special right triangles is that by using the Pythagorean theorem, we can derive the sides of both of the special right triangles. I can understand how the values came to be in the special right triangles, such as how the 60 degree side of the 30-60-90 triangle was originally radical 3 over 2, until we simplified the sides of the triangle further. It is amazing how the Pythagorean triangle can be applied to so many things!
Being able to derive these patterns myself aids in my learning because if I ever forget the rules of the 45-45-90 or 30-60-90 triangle, I will be able to derive them from the square and the equilateral triangle with the sides lengths of 1. I will know how to use the Pythagorean theorem to determine the sides and then substitute the values for the variable n, so I know that any other multiples of those values can be used. It broadens my understanding of the special right triangles even further because now I have a better grasp of where they come from and what different math concepts could be applied in deriving the special right triangles.