Calculating an approximation of pi using a square inside a circle

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Calculating an approximation of pi using a square inside a circle

Mathematics
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|dw:1441563540851:dw|
I need to find a close approximation of pi using a square inside a circle
|dw:1441563708086:dw| well you could say \[\text{ small square area }<\text{ circle area } < \text{ big square area }\] |dw:1441563787054:dw| |dw:1441563837206:dw| so you have: \[r^2 <\text{ circle area}

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Ok. I also know that the radius is 1. Let me calculate the areas
well I didn't know if we were just limited to using the square inside the circle..
Well..how would you do it with two squares?
I drew two squares above..
what do you mean?
Sorry, I explained myself wrong. Give me a minute, please
oops and I wrote things wrong aboe
\[(2r)^2< \text{ circle area } < (2R^2) \\ \text{ circle area } \approx \frac{(2r)^2+(2R)^2}{2}\]
|dw:1441564320576:dw||dw:1441564336095:dw|
I used R and r as half the side lengths above
Yep. So we know that R is 1, which should make things a little easier. How can we calculate r?
Right now I have pi = 2r^2 + 2
|dw:1441564624855:dw|
r is sqrt(2)/2
\[2(\frac{\sqrt{2}}{2})^2+2 =2(\frac{2}{4})+2=1+2=3\]
Great! Thanks a lot. And any idea of how to do this with 1 square? (just for curiosity)
well I guess you can get a lower approximation of the circle just by consider the area of the square
area of the small square*
Ok, thanks a lot :) Have a nice day
you too
it kind takes me back to calculus days
we can find the lower sum and the upper sum but I think the average of those sums gave a better approximation
They told me something about finding the max square that could be inside the circle
we can find the lower sum and the upper sum but I think the average of those sums gave a better approximation
Ok,thanks again :)

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