1. **Problem statement:** We have four rectangles A, B, C, and D, all with the same perimeter. We need to order them from greatest to least area. 2. **Formula and rules:** The perimeter $P$ of a rectangle with length $l$ and width $w$ is given by: $$P = 2(l + w)$$ The area $A$ is: $$A = l \times w$$ For a fixed perimeter, the rectangle with the greatest area is a square. This is because for a fixed sum $l + w$, the product $l \times w$ is maximized when $l = w$. 3. **Analysis:** Since all rectangles have the same perimeter, their $l + w$ is constant. The area depends on how close $l$ and $w$ are to each other. - Rectangle A is horizontally oriented, so $l > w$ but closer to a square. - Rectangle B is narrow and vertical, so $w$ is much smaller than $l$. - Rectangle C is long horizontally, so $l$ is much larger than $w$. - Rectangle D is taller vertical, so $w$ is larger but still not equal to $l$. 4. **Ordering areas:** The closer the sides are to equal, the larger the area. So the order from greatest to least area is: $$A > D > B > C$$ 5. **Second question:** Is it possible to draw a rectangle with the same perimeter but greater area? Yes. The rectangle with the greatest area for a given perimeter is a square. If none of the given rectangles is a square, then a square with the same perimeter will have a greater area. 6. **Explanation:** For perimeter $P$, the square side length is: $$s = \frac{P}{4}$$ Area of the square: $$A_{square} = s^2 = \left(\frac{P}{4}\right)^2 = \frac{P^2}{16}$$ This area is greater than any rectangle with the same perimeter but unequal sides. 7. **Diagram:** Imagine adjusting the sides of any rectangle to become more equal while keeping the perimeter constant; the area increases until it reaches the maximum at the square. **Final answers:** 1. Order from greatest to least area: $A > D > B > C$ 2. Yes, a square with the same perimeter has the greatest area among rectangles.