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

INQUIRY ACTIVITY SUMMARY:

In order to derive the pattern for a 45-45-90 triangle from from a square with a side length of 1, we first need to cut the square from corner to corner, as shown in the picture, to get the triangle. Since the square has a side length of 1, we know that the x and y both also equal one so we can label the sides 1.

Now you have the sides, so all you need no is the hypotenuse. To find it we will use the pythagorean theorem. a^2+b^2=c^2. plug in the numbers we have. Since we know a and b both equal 1 then we use 1+1=c^2 since 1^2 still equals 1. We end up having C^2=2. so we have to cancel he square by finding the square root of c^2 which will equal C... But what you do to one side, you do to the other so it'll end up being the square root of 2.

Now that we have the  numbers, we will now add the variable because the sides are not always gonna equal 1. You can use any letter for the variable, i used X. Since both a and b both equal 1 we can just use X for both of those sides. To find the hypotenuse we have to use the pythagorean theorem. we plug in X^2+X^2= 2X^2

Now that we're done with the 45-45-90, we can move on to the 30-60-90 from an equilateral triangle with a side length of 1 and angles of 60-60-60. We start by cutting it right down the middle.

 Since we cut it right down the middle. the top corner is also cut in half so instead of a 60, its now 2 30's. And also the bottom is cut in half so instead of being 1 is it 2 .5's or 1/2's.

We, again, have to use Pythagorean theorem to find the the Y or b. First we have to manipulate a (1/2)^2+b^2=1. Once we do, it will end up being b^2=1-(1/4). That means b will equal to radical3/2.

since these are just for the side, the letter variables you use are the exact same thing. Hypotenuse=n

x=n/2 y=nradical3/2. To get rid of the fractions all you do is multiply each side by 2. Hyp.=2n  x=n  y=n radical3. N is just what you use when The hypotenuse is unknown. so the completed triangle should look like this.

INQUIRY ACTIVITY REFLECTION

Something I never noticed before about special right triangles they are the exact same things as the Unit circle.

Being able to derive these patterns myself aids in my learning because it makes me independent and i don't have to rely on anything because i have all this in my head.