I/D #2: Unit O- Deriving the SRTs

INQUIRY ACTIVITY SUMMARY:

1.) 30-60-90

(Deriving a 30-60-90 triangle from an equilateral triangle)

In a 30-60-90 triangle, we know that it is derived from an equilateral triangle. In an equilateral triangle, all the angles add up to 180, 60 at each angle. Since we know that the sides are the same in an equilateral, we can label them all the same. We are given 1 as our side length in this activity. Once we split the triangle in half, we get a 30-60-90 triangle. The value of a is 1/2 since it was split from 1 and c (the hypotenuse) stays the same as 1. To solve for the last side, b, we must apply the Pythagorean Theroem.

(finding the missing vairable, b using the Pythagorean Theorem)

We know that the Pythagorean Theorem is a^2+b^2=c^2. If we plug in our known variables, which will be a and c, we can find the missing variable, b. After we plug it in and solve, we are left with b=(rad)3/2 after finding the squareroot of (rad)3/4.

(Triangle after all sides are found)

Once we figure out the side length for each side, we are left with the above picture. Dealing with fractions can be such a hassle so it's easier to just multiply everything by 2 and get a non-fraction answer for each side. We are left with a=1, b=(rad)3, and c=2.

(final 30-60-90 triangle derived from an equilateral triangle)

When we are using SRT's it doesn't necesarily always have to have a hypotenuse of 1. By putting the variable n where 1 would be, it allows us to plug in any given value and keep the ratio at the same time. So in the above picture, we are left with the variables n(rad)3, 2n, and n. When given a value for n, we can easily plug it in and figure out the value of each side.

2.) 45-45-90

(Deriving a 45-45-90 triangle from a square)

When deriving a 45-45-90 triangle from a square with a side length of one, we know that the sides will be 1 all around. We know that in a square, all the angles are 90 because the total degree of a square is Once we split the square in half, our angles will be 45-45-90. Our a and b value will be 1 and to solve for side, c (the hypotenuse), we can apply the Pythagorean Theorem.

(Using Pythagorean Theorem to solve for missing variable, c)

To find the missing sides, c, we plug in our known variables into the Pythagorean Theorem. We end up with c=(rad)2. 

(45-45-90 triangle with all the sides found)

After we find the missing piece, our 45-45-90 triangle will end up like this. Unlike the 30-60-90 triangle where we had to deal with fractions, we can just leave the numbers alone since they're whole numbers.

(final 45-45-90 triangle derived from a square)

After finding all the sides, we can do the same thing as in the previous trianlgle and plug n into where the 1's are. We are left with n(rad)2, n and n. The n is used to plug in any given value while keeping the ratio, which represents the relationshp between the three sides of a special right triangle.

INQUIRY ACTIVITY REFLECTION

1. Something I never noticed before about special right triangles is how it derived from a square (for a 45-45-90 triangle) or an equilateral triangle (for a 30-60-90 triangle).

2. Being able to depreive these patterns myself aids in my learning because now I can derive it on my own and I know where the values came to be.