1. **State the problem:** We need to find the volume of a solid composed of a right pyramid mounted on a rectangular block. Both have a square base of side length 6 cm. The block height is 10 cm, and the pyramid's apex V is such that VA = VB = VC = VD = 5 cm. 2. **Calculate the volume of the rectangular block:** The block is a rectangular prism with base area $6 \times 6 = 36$ cm$^2$ and height 10 cm. $$\text{Volume}_{\text{block}} = \text{base area} \times \text{height} = 36 \times 10 = 360 \text{ cm}^3$$ 3. **Calculate the height of the pyramid:** The pyramid has a square base of side 6 cm and apex V such that the distances from V to each vertex of the base (A, B, C, D) are 5 cm. The center of the square base is the midpoint of the square, call it O. The distance from O to any vertex (e.g., A) is half the diagonal of the square: $$OA = \frac{\sqrt{6^2 + 6^2}}{2} = \frac{\sqrt{72}}{2} = \frac{6\sqrt{2}}{2} = 3\sqrt{2} \approx 4.2426 \text{ cm}$$ Since VA = 5 cm and V lies above O, the height $h$ of the pyramid is found by the Pythagorean theorem in triangle VOA: $$VA^2 = VO^2 + OA^2 \implies VO^2 = VA^2 - OA^2 = 5^2 - (3\sqrt{2})^2 = 25 - 18 = 7$$ $$h = VO = \sqrt{7} \approx 2.6458 \text{ cm}$$ 4. **Calculate the volume of the pyramid:** The volume of a pyramid is given by: $$\text{Volume}_{\text{pyramid}} = \frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 36 \times \sqrt{7} = 12 \sqrt{7} \approx 31.75 \text{ cm}^3$$ 5. **Calculate the total volume of the solid:** $$\text{Volume}_{\text{solid}} = \text{Volume}_{\text{block}} + \text{Volume}_{\text{pyramid}} = 360 + 12 \sqrt{7} \approx 360 + 31.75 = 391.75 \text{ cm}^3$$ **Final answer:** The volume of the solid is approximately $391.75$ cm$^3$.