1. The problem involves calculating the area and perimeter of squares and rectangles given their side lengths or dimensions. 2. For a square, the area $A$ is given by the formula $$A = s^2$$ where $s$ is the length of a side. 3. The perimeter $P$ of a square is given by $$P = 4s$$. 4. For a rectangle, the area $A$ is $$A = l \times w$$ where $l$ is the length and $w$ is the width. 5. The perimeter $P$ of a rectangle is $$P = 2(l + w)$$. 6. Let's calculate for each figure: **Square with $s=56$ ft:** - Area: $$A = 56^2 = 3136\text{ ft}^2$$ - Perimeter: $$P = 4 \times 56 = 224\text{ ft}$$ **Square with $s=52$ inches:** - Area: $$A = 52^2 = 2704\text{ in}^2$$ - Perimeter: $$P = 4 \times 52 = 208\text{ in}$$ **Square with $s=67$ ft:** - Area: $$A = 67^2 = 4489\text{ ft}^2$$ - Perimeter: $$P = 4 \times 67 = 268\text{ ft}$$ **Square with $s=70$ ft:** - Area: $$A = 70^2 = 4900\text{ ft}^2$$ - Perimeter: $$P = 4 \times 70 = 280\text{ ft}$$ **Rectangle with $l=78$ inches, $w=60$ inches:** - Area: $$A = 78 \times 60 = 4680\text{ in}^2$$ - Perimeter: $$P = 2(78 + 60) = 2 \times 138 = 276\text{ in}$$ **Rectangle with $l=84$ ft, $w=50$ ft:** - Area: $$A = 84 \times 50 = 4200\text{ ft}^2$$ - Perimeter: $$P = 2(84 + 50) = 2 \times 134 = 268\text{ ft}$$ **Square with $s=50$ inches:** - Area: $$A = 50^2 = 2500\text{ in}^2$$ - Perimeter: $$P = 4 \times 50 = 200\text{ in}$$ **Rectangle with $l=83$ inches, $w=54$ inches:** - Area: $$A = 83 \times 54 = 4482\text{ in}^2$$ - Perimeter: $$P = 2(83 + 54) = 2 \times 137 = 274\text{ in}$$ **Rectangle with $l=76$ inches, $w=41$ inches:** - Area: $$A = 76 \times 41 = 3116\text{ in}^2$$ - Perimeter: $$P = 2(76 + 41) = 2 \times 117 = 234\text{ in}$$ These calculations show how to find the area and perimeter for each shape using the given side lengths.