1. **Problem statement:** Find the volume of each pyramid given the dimensions. 2. **Formula:** The volume $V$ of a pyramid is given by $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ 3. **Important rules:** - For triangular bases, Base Area = $\frac{1}{2} \times \text{base edge}_1 \times \text{base edge}_2$ (assuming right triangle or given perpendicular edges). - For square bases, Base Area = side$^2$. - Always multiply base area by height and then divide by 3. --- ### Problem 30 - Base edges: 2.4 in and 1.9 in (triangular base) - Height: 2.3 in Calculate base area: $$\text{Base Area} = \frac{1}{2} \times 2.4 \times 1.9 = \frac{1}{2} \times 4.56 = 2.28$$ Calculate volume: $$V = \frac{1}{3} \times 2.28 \times 2.3 = \frac{1}{3} \times 5.244 = \cancel{\frac{1}{3}} \times 5.244 = 1.748$$ Rounded to nearest tenth: $$V \approx 1.7 \text{ in}^3$$ --- ### Problem 31 - Base edges: 12 ft and 5 ft (triangular base) - Height: 11 ft Calculate base area: $$\text{Base Area} = \frac{1}{2} \times 12 \times 5 = \frac{1}{2} \times 60 = 30$$ Calculate volume: $$V = \frac{1}{3} \times 30 \times 11 = \frac{1}{3} \times 330 = 110$$ --- ### Problem 32 - Base edges: 13.1 m and 9.2 m (triangular base) - Height: 12 m Calculate base area: $$\text{Base Area} = \frac{1}{2} \times 13.1 \times 9.2 = \frac{1}{2} \times 120.52 = 60.26$$ Calculate volume: $$V = \frac{1}{3} \times 60.26 \times 12 = \frac{1}{3} \times 723.12 = 241.04$$ Rounded to nearest tenth: $$V \approx 241.0 \text{ m}^3$$ --- ### Problem 33 - Square base edges: 6 cm each - Height: 7.5 cm Calculate base area: $$\text{Base Area} = 6^2 = 36$$ Calculate volume: $$V = \frac{1}{3} \times 36 \times 7.5 = \frac{1}{3} \times 270 = 90$$ --- **Final answers:** - 30: $1.7$ in³ - 31: $110$ ft³ - 32: $241.0$ m³ - 33: $90$ cm³