Tuesday, March 4, 2014

ID#2 Unit O Concept 7-8

So basically in this paper we are going to derive an equilateral square and triangle. With these we are going to get a 45-45-90 triangle and the 30-60-90 triangle. when we cut the in half we will get these special right triangles and for us to solve the missing side we are logically going to use the Pythagorean  theorem.
45-45-90 Triangle

In this picture you can see what we are starting off with, it is an equilateral square.

Here we can see that we already cut it in have so since all sides are 90 when you cut it in half your other angles are going to be 45 degrees.

Here I am doing the work for the use of the Pythagorean theorem. we will use a^2+ b^2 = c^2 since we already have both sides it is obviously one and since we have to take the radical to get c alone it will end up being radical 2 for out hypotenuse answer.
Here we can see that the both sides will always be the same number so we can go ahead and substitute it got "n" and then for the hypotenuse we will always have to multiply with radical 2 so for example we have one side of the 45 degree angle 5 the other side must me 5 and the hypotenuse 5 radical 2.

  30-60-90- Triangle

Here we can see how we have our equilateral triangle

now here we have already but the triangle in half and we have all the degrees there too. all sides on an equilateral triangle are 60 degrees and when you cut it in half we will have 30 and 30 degrees on the sides.

So when we have to do the Pythagorean theorem we can see that the bottom will have to split into 2 so one divided by 2 is one half. when we have that we can plug everything in. we edn up getting radical 3 over 2 for side b.

Here we can see that we really don't want to use the radial 3 over 2 because it will make it difficult so we are going to multiply everything by 2 to make it easier when we do we are going to add an n to represent any number so when we have that we and up getting one for side a radial 3 for side b and 2 for the hypotenuse but got this triangle there are different degrees so the 60 side as you can see has the radical 3 so it will always have to do with that. when we have n equal to 6 side b will be 6 radial 3 and the hypotenuse will be 12. when we do the problem backward for example and b equals 8 we will have to go back to the a side by dividing by radial 3 which will give us 8 rad 3 over 3 and the hypotenuse 16 rad 3 over 3 .


1. Something I never noticed before about special right triangles was that they all had the same ration between each other and how we found that out.
2. Being able to derive these patters myself aids in my learning because i know how i got there and how those triangle were made


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