3 ms·
Her rectangle problem is wrong. Here are four rectangles with the same area and perimeter: x = 3, y = 6; x = 4, y = 4; x = 5, y = 10/3; x = 7, y = 14/5 The f
by robertk 17y ago
Her rectangle problem is wrong. Here are four rectangles with the same area and perimeter:
x = 3, y = 6; x = 4, y = 4; x = 5, y = 10/3; x = 7, y = 14/5
The formula is fix x > 2 (one side), let y = 2x / (x-2) (other side). Derived from xy = 2x + 2y.
- gort 17y agoAs I recall the problem was not "there aren't any" but "there aren't any more"