1. **Problem statement:** We have a pyramid with a square base ABCD where each side is $\sqrt{72}$ cm. The diagonals AC and BD intersect at point P. The vertex V is vertically above P with height VP = 8 cm. We need to find: (a) The volume of the pyramid. (b) The length of diagonal AC. 2. **Formulas and rules:** - Volume of a pyramid: $$V = \frac{1}{3} \times \text{Base Area} \times \text{Height}$$ - Area of a square: $$\text{side}^2$$ - Length of diagonal of a square: $$\text{diagonal} = \text{side} \times \sqrt{2}$$ 3. **Find the base area:** Side length = $\sqrt{72}$ cm Base area = $\left(\sqrt{72}\right)^2 = 72$ cm$^2$ 4. **Find the volume:** Height = VP = 8 cm Volume = $\frac{1}{3} \times 72 \times 8 = \frac{1}{3} \times 576 = 192$ cm$^3$ 5. **Find the length AC:** Diagonal AC = side $\times \sqrt{2} = \sqrt{72} \times \sqrt{2} = \sqrt{72 \times 2} = \sqrt{144} = 12$ cm **Final answers:** (a) Volume = 192 cm$^3$ (b) AC = 12 cm