I/D2: Unit O- How can we derive the patterns for our special right triangles?

Inquiry Activity Summary

30-60-90 Triangles

Before we can solve a 30-60-90 triangle, we need to know how to get 30-60-90 triangles. On this worksheet we are given a 60-60-60 triangle (equiangular), and in order to get a 30-60-90, we must cut down the middle of it, giving us 2 triangles, both 30-60-90. To derive the patterns for 30-60-90 triangles, we must first know what to use to solve for each side. One way to get the side measures is using a special pattern. This pattern says that the shortest side is 'n', the hypotenuse is '2n', and the height is 'n√3'. But because this triangle is a right triangle, we can also use the Pythagorean theorem. But we need two sides in order to solve for one missing side; unfortunately we are only given one side measure: the hypotenuse (equals 1), that isn't an issue in this case because in these specific triangles, the shortest side is always half the size of the hypotenuse. Now we have the size of 2 sides; now we can use the Pythagorean theorem: a²+b²=c² . For this example, I'll label the short leg "b".  a²+(⅟₂)² =1² so simplified, a= ^√3/_2. 

45-45-90 Triangles

Deriving the pattern for a 45-45-90 triangle is simple and somewhat similar to the previous triangle. In order to get a 45-45-90 triangle, we must split the square in half diagonally, we do so because each angle in a square is 90˚ and splitting the opposites in half gives us 45˚ on both sides; leaving us with a 45-45-90 triangle. Because this triangle is also a right triangle, we can also use the Pythagorean theorem to solve for missing sides. In this activity we have the side lengths. For any of these triangles, "a" and "b" are the same lengths, and because of this fact about 45-45-90 triangles, filling in the Pythagorean formula is easier. 1²+1²=c² is the result of substituting in the formula, fully simplified: c=√2.

INQUIRY ACTIVITY REFLECTION

Something I never noticed about special right triangles is... that the easiest way to solve for sides is to use the Pythagorean theorem.

Being able to derive these patterns myself aids in my learning because... it gave me a visual representation which helped me better my understanding of special right triangles.