osanseviero
  • osanseviero
Get an approximation of pi using a square and triangles inside a circle.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
osanseviero
  • osanseviero
|dw:1441838480540:dw|
osanseviero
  • osanseviero
The idea is to first calculate the area of the square and then add triangles.
osanseviero
  • osanseviero
|dw:1441838520139:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

osanseviero
  • osanseviero
Asquare = (2r)^2 = 2 My problem is with getting the first level of triangles, and then adding triangles to those triangles
phi
  • phi
what did you pick for r ? how about starting with r= 1?
osanseviero
  • osanseviero
I'm using r for distance from center to square. The radius is 1.
osanseviero
  • osanseviero
So with that I can get a height for a triangle 1-r.
phi
  • phi
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
osanseviero
  • osanseviero
Ok, lets use x instead.
phi
  • phi
the height of the triangle will be r- x 1 - \sqr(2)/2
osanseviero
  • osanseviero
Ok, and base would be x*2,right?
phi
  • phi
yes
osanseviero
  • osanseviero
Ok. So that would be ((1 - sqrt(2) / 2)(sqrt(2))/2. That times 4 to get total area
phi
  • phi
yes
osanseviero
  • osanseviero
Ok, I was good until here. I started to get lost when adding a new layer of triangles
osanseviero
  • osanseviero
|dw:1441839343074:dw|
osanseviero
  • osanseviero
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
phi
  • phi
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
osanseviero
  • osanseviero
Ok, give me a minute, I'll try to get it
osanseviero
  • osanseviero
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)
osanseviero
  • osanseviero
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|
osanseviero
  • osanseviero
Oh...that would be the previous base/2
anonymous
  • anonymous
in my mind it looks like this
anonymous
  • anonymous
|dw:1441840270963:dw|
phi
  • phi
this looks like a painful process
osanseviero
  • osanseviero
It is haha. I think that with this we should get a really close approximation
phi
  • phi
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
osanseviero
  • osanseviero
It is easier, I already did that one hehe
osanseviero
  • osanseviero
Thanks for the help :)
osanseviero
  • osanseviero
Can you just help me get that last height?
phi
  • phi
it will take some time
osanseviero
  • osanseviero
Ok, then don't worry :)
phi
  • phi
|dw:1441843190370:dw| find "y" using pythagoras and the altitude is 1-y

Looking for something else?

Not the answer you are looking for? Search for more explanations.