The idea is to first calculate the area of the square and then add triangles.
Asquare = (2r)^2 = 2 My problem is with getting the first level of triangles, and then adding triangles to those triangles
what did you pick for r ? how about starting with r= 1?
I'm using r for distance from center to square. The radius is 1.
So with that I can get a height for a triangle 1-r.
ok, but using r for the length of 1/2 the side of the square is confusing (people see "r" and think radius) can I use x instead of r? diameter is 2, and this is the length of the diagonal of the square so its side is 2/sqr(2) = sqr(2) and x= 1/2 of a side = sqr(2)/2
Ok, lets use x instead.
the height of the triangle will be r- x 1 - \sqr(2)/2
Ok, and base would be x*2,right?
Ok. So that would be ((1 - sqrt(2) / 2)(sqrt(2))/2. That times 4 to get total area
Ok, I was good until here. I started to get lost when adding a new layer of triangles
So I guess it will pass through the half of the half of the base of the triangle, and then half of the side of the triangle
yes, but now it's not clear what the altitude of the new triangle is. the base of the new triangle is the "hypotenuse" of the old triangle, so we can find it using pythagoras
Ok, give me a minute, I'll try to get it
Ok. The base is pretty simple. The altitude is a little harder. First, I would calculate the distance between the center and the intersection with the square. Then add the intersection with the triangle. For the square, I did: |dw:1441839701196:dw| So a = sqrt((x/2)^2 + x^2)
How to find where the line touches the hypotenuse of the second triangle? We know that the height would be the half.|dw:1441839908971:dw|
Oh...that would be the previous base/2
in my mind it looks like this
this looks like a painful process
It is haha. I think that with this we should get a really close approximation
I know archimedes found the area of triangles formed by an n-gon (n sided polygon) on the inside of the circle and the outside and so the area of the circle is between those two areas. But that process seems easier than adding smaller and smaller triangles
It is easier, I already did that one hehe
Thanks for the help :)
Can you just help me get that last height?
it will take some time
Ok, then don't worry :)
|dw:1441843190370:dw| find "y" using pythagoras and the altitude is 1-y