Tuesday, March 4, 2014

I/D#2: Unit O Concept 7-8: Deriving Patterns for Special Right Triangles

Inquiry Activity Summary:

The purpose of this activity was to help us understand why special right triangles have the patterns they do. For example, a 45-45-90 triangle has side lengths of n, n, and n√2 while a 30-60-90 triangle has side lengths of n, n√3, and 2n. Instead of memorizing two seemingly random patterns, we are now able to understand where these patterns come from so if we forget the patterns, we can easily figure it out ourselves, since we now know how to derive them.

1) 30°-60°-90° Triangle

We can construct a 30°-60°-90° triangle bu cutting an equilateral triangle of any side length (in this case to keep it simple, is 1) down its altitude to make 2 right triangles. As we all know equilateral triangles have equal sides and equal angles(which are 60°) and because we cut one angle and one side in half, there is now two 30° angles and the side length of the smaller triangles is 1/2. So far, we know two sides of the smaller triangle: hypotenuse of 1 and one side(opposite 30° angle) length of 1/2. We use Pythagorean theorem to solve for the last side to end up with √3/2. Now that we know all the sides and angles of the
30°-60°-90° triangle, we can start to see the relationship between the sides and how we got the pattern. The hypotenuse is 2 times the length of the shortest leg of 1/2 while the longer leg looks to √3 times 1/2. Therefore, if we multiply n to each of the sides and multiply by 2(to make the numbers easier), we get sides of n, n√3, and 2n. This applies to any number n might equal to. The pattern is still the same because n is just a constant.

2) 45°-45°-90° Triangle

We construct 45°-45°-90° triangle by cutting a square (side lengths 1 to keep it simple) down its diagonal, cutting the 90° angles in half and making two 45°-45°-90° triangles. Because the triangle still has 2 sides of the original square, we know their values are both 1. Then, we use the Pythagorean theorem to solve of the length of the hypotenuse, which in this case is just √2. Looking at our triangle with all of its sides and angles solve for, we can see the relationship of the sides already. The two legs are equal with a length of 1 while the hypotenuse is 1 times √2. Now if we multiply n to all 3 sides, we end up with sides of n, n, and n√2, making the pattern. N is again is just a constant that applies to any side length because the relationship between the sides is still the same. That way it is easier to remember the pattern rather than memorizing sets of numbers that apply this pattern.

Inquiry Activity Reflection:
1. "Something I never noticed before about special right triangles is..." that there patterns can be logically derived from a process involving Pythagorean theorem and cutting squares/triangles.
2. "Being able to derive these patterns myself aids in my learning because..." I can now easily and quickly calculate the side lengths of the SRTs. That means I don't have to memorize sets of numbers that could apply to these triangles or the pattern because if I forget the pattern, I can easily derive it again and use it to find the sides, instead of remembering a random pattern.