1. **Square with diagonal $3\sqrt{2}$:** The formula relating the diagonal $d$ and side length $s$ of a square is $$d = s\sqrt{2}$$ Given $d = 3\sqrt{2}$, solve for $s$: $$3\sqrt{2} = s\sqrt{2}$$ Divide both sides by $\sqrt{2}$: $$\cancel{\sqrt{2}} \times 3 = s \cancel{\sqrt{2}}$$ $$s = 3$$ Area of square = $s^2 = 3^2 = 9$ square units. 2. **Circle with diameter 24 ft:** Radius $r = \frac{d}{2} = \frac{24}{2} = 12$ ft. Area of circle = $$\pi r^2 = \pi \times 12^2 = 144\pi$$ square feet. 3. **Shape with semicircular ends and rectangle:** Rectangle area = length $\times$ height = $30 \times 15 = 450$ cm$^2$. Radius of semicircle = half the height = $\frac{15}{2} = 7.5$ cm. Area of one semicircle = $$\frac{1}{2} \pi r^2 = \frac{1}{2} \pi (7.5)^2 = \frac{1}{2} \pi \times 56.25 = 28.125\pi$$ cm$^2$. Two semicircles make a full circle, so total semicircle area = $2 \times 28.125\pi = 56.25\pi$ cm$^2$. Total area = rectangle area + semicircle area = $450 + 56.25\pi$ cm$^2$. 4. **Polygon with two right triangles on top of a rectangle:** Rectangle area = width $\times$ height = $30 \times 15 = 450$ in$^2$. First triangle base = 15 in., height = 10 in. Area = $$\frac{1}{2} \times 15 \times 10 = 75$$ in$^2$. Second triangle base = 5 in., height = 10 in. Area = $$\frac{1}{2} \times 5 \times 10 = 25$$ in$^2$. Total area = rectangle + both triangles = $450 + 75 + 25 = 550$ in$^2$. **Final answers:** - Square area = $9$ units$^2$ - Circle area = $144\pi$ ft$^2$ - Shape area = $450 + 56.25\pi$ cm$^2$ - Polygon area = $550$ in$^2$