I don't think that anyone can tell you precisely how to do it.  You must have a decent imagination and a good understanding of geometry.  With that said:

    The top circle is in the corner of the square, therfore the radius of the circle is one inch away from each side.  Using the Pythagorean Theorem (x^2 + y^2 = r^2) tells you that the center of the first circle is 2^(1/2) away from the corner.  This relationship will hold in the opposite corner as well.

     So the total diagonal distance will be those two distances from the opposite corners plus the entire diameter of the center circle plus the two half circle or radius remaiders from the corner circles.

     The diagonal distance would then be: 2(2)^(1/2) + 2 + 2

     The sides of the square will also be solved using the Pythagorean Theorem with the diagonal distance being the hypotenuse.

     x^2 + y^2 = r^2,   since x = y in this case,  2x^2 = r^2

     Our radius from earlier was 4 + 2(2)^(1/2) = 4 + 8^(1/2)

    2x^2 = 4 + 8^(1/2)

    x^2 = (4 + 8^(1/2)) / 2

    x = (2 + 2^(1/2))^(1/2)

    Since the area of a square is simply a side squared the area is:

    A = 2 + 2^(1/2)

That's the exact answer.  It is approximately 3.41421356237.................

Last edited by irspow (2005-12-08 10:13:13)

wait whats the answer??