Consider a rectangle with sides of length a ≤ b inside the square. Since b≤1 and 2a+2b= 5/2 hold,we see that a ≥ 1/4. The area of the rectangle equals ab and is therefore at least 0.25 × 0.25 = 1 . Hence, we can have no more than 16 rectangles inside the square without creating overlaps. ( b) A solution is sketched in the figure below. The four outer rectangles, A through D, are equal with the shorter side having length x, and the longer side having length 1 − x. Together they leave uncovered a square area with sides of length 1 − 2x. This area is then tiled by 26 equal rectangles. These have sides of length 1 − 2x and (1−2x)/26, and therefore have a circumference of (54/26)(1 − 2x). To obtain a circumference of length 2,we take x = 1/54