ID#2 Unit O Concept 7-8: How can we derive the patterns for our special right triangles?

Inquiry Activity Summary

       In the activity that we were given in class we were supposed to derive a 45-45-90 out of a square making sure that both of the triangles that we make produce a 45-45-90 triangle.  All of the sides of the square were equaling to 1 and we had to find the value of the sided and the hypotenuse, making sure that it will still be equal to 1. Then to form the 30-60-90 we were given an equilateral triangle in which all the sides were also equaling 1. For the 30-60-90 we also had to find the hypotenuse and the two sides all equaling to 1 at the end.

45-45-90

       For the 45-45-90 we are given a square in which obviously all sides are equal but they are equal to 1 and we have to make a 45-45-90 triangle out of it. I found the 45-45-90 triangle by dividing the square diagonally to split the 90 degrees into 45 degrees leaving the triangle with only one 90 and two 45 degree angles. From there we have to find out how to make all the sides still equal to 1 even when it has been divided. what is missing this time is the hypotenuse which means that you have to do the Pythagorean to find the hypotenuse and since each side is still 1 all you have to do is square 1 which is still one and add them which will equal to 2 and from there you have to square root it and that will be rad. 2. the "n' was being used to be able to make the triangles in different sizes using "n" as the ratio in which they can be interchangeable.

30-60-90

      For the 30-60-90 we were given an equilateral triangle in which had to set up the triangle to be able to make two 30-60-90 triangles to have all sides still equal 1. I divided the triangle straight down the middle to form the two triangles, if you divide it down the middle you get the 30 degrees at the top since you split 60 in half, the 60 at the side, and the 90 down the middle since you divided 180 by two. Once you have divided it by the middle you now have the sides equal 1 and the bottom both equal 1/2. To be able to get the variables to equal to "n", "2n", and "n rad. 3" you have to do the Pythagorean to find you b since you already have your hypotenuse and one side you plug in what you know. once you have completed the Pythagorean you will know the sides but since it is in fractions you can make it into whole numbers by multiplying my two to get rid of the 2 at the bottom. Using "n" allows for there to be a ratio in the equation therefore being able to enlarge or to shrink the triangle.

Inquiry Activity Reflection

Something i never noticed before about special right triangles is that they derive from a square or an equilateral triangle and that how the variable derive from.

Being able to derive these patterns myself aids in my learning because now i further understand where the equations come from, they make sense as into form where the "n" comes from and why it has the radicals or other numbers in front of them.