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And, it looks like imatfaal and I came up with the same answer []

If you are thinking of a square that precisely fits into a circle.Or a circle that precisely fits into a square.[diagram=627_0]Then you can calculate the areas of each.Using r as the radius of the circle.Square in Circle:Area of circle: πr^{2}Area of the square. = 4 x triangle with base&height equal to r4 x r^{2}/2 = 2r^{2}Difference is: πr^{2} - 2r^{2} = (π-2)r^{2} = (3.14-2)r^{2} = 1.14r^{2}Circle in Square:Area of circle: πr^{2}Area of square = 4 x squares with base&height equal to r4r^{2}Difference is:4r^{2}-πr^{2} = (4-π)r^{2} = (4-3.14)r^{2} = 0.86r^{2}But..As the previous poster (imatfaal) mentioned, taking differences isn't adequate as the areas are different.So...Taking a ratio of the inner to the outer is probably a better way to look at it.Square in Circle:2r^{2}/πr^{2} = 2/π = 2/3.14 = 0.64Circle in Square:πr^{2}/4r^{2} = π/4 = 3.14/4 = 0.785So the circle in the square actually wins with more area in the square vs the area in the outer container.And, it looks like imatfaal and I came up with the same answer []

great piece i gotta read again but more urgently , i should try to beat a square peg in a round hole instead of a round peg in a square hole?

in the time you spent to quote and write above - why not read either of the answers given?

Quote from: CZARCAR on 23/05/2011 20:44:50great piece i gotta read again but more urgently , i should try to beat a square peg in a round hole instead of a round peg in a square hole?in the time you spent to quote and write above - why not read either of the answers given? edited to fix quotes