Relationship between circle areas. Can anyone give me a text reference for this proof? See details.?

Question

The area of a circle drawn through the 4 points of a square is exactly twice the area of the circle inscribed within the square and touching the squares sides. This geometrical fact must be registered in some text. I have searched the net, including Mathworld, but cannot find a reference to such a proof. Can anyone give me a reference?

Answer 1

Here's something similar that you can work with:
http://www.cs.miami.edu/~burt/manuscripts/archimedes/ad_I_I.pdf
or this: http://web.pdx.edu/~dgreene/circlesandsquares.html

Answer 2

I don't have any reference, but i can explain this to you why this is true

area of a circle = Pi x r^2

When you draw a circle Around Outside a square, you are touching the corners of the square. Distance from one corner to the opposite corner is the diagonal of the square

Using Pythagorean theorem, IF we have square with side lenght of 'a', our diagonal will measure: SquareRoot(2) x a

Now this diagonal is the diameter of the circle outside the square.

Radius is half diameter, so we have Radius of big circle = SquareRoot(2) x a/2


Area of the Big circle is: Pi x [squareRoot(2) x a/2 ]^2

When you perform that square, you get rid of the square root and are left with: Pi x 2 x [a/2]^2 = Pi x 2 x a^2 / 4 = Pi x a^2 /2

Now in case of the inner circle:

The inner circle touches accross the square

so the diameter is just equal to: a

Radius of smaller circle is a/2

Area of smaller circle is: Pi x [a/2]^2

Area = Pi x a^2 / 4

Now you can see that the area of the smaller circle is half the bigger circle

Answer 3

If the length of one side of a square is s, then by using the Pythagorean theorem you can show that the length of the square's diagonal is (square root of 2)*s.

The radius of the inscribed circle is half the length of the square's side = s/2, so its area a = (pi)((s/2)^2) = ((pi)(s^2))/4.

The radius of the circumscribed circle is half the length of the square's diagonal = ((square root of 2)s) / 2, so its area A = ((pi)((square root of 2)s) / 2)^2
= (pi)((square root of 2)^2)(s^2)) / 4
= ((pi)(2)(s^2)) / 4
= ((pi)(s^2)) / 2
= 2a

This would probably not be proven as a theorem in any geometry textbook, although it might be given as an exercise.

Answer 4

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