In a square of side length 40, can we fit a 10 x 45 rectangle?
Note: You cannot fold / break the rectangle or the square.
+128475 As EP pointed out, it will fit diagonally...
Let the bottom left vertex of the square be located at (0,0)
And the two vertexes of the short side of the rectangle can be located at (0, 10/√2) and ( 10/√2, 0)
And let the other two vertexes be located at (40 - 10/√2, 40) and ( 40, 40 - 10/√2)
And the distance between (10/√2, 0 ) and ( 40 , 40 - 10/√2) =
√ [ (40 - 10/√2)^2 + (40 - 10/√2)^2 ] = √2 * (40 - 10/√2) = 40√2 - 10 ≈ 46.56
So.....the rectangle will fit
