1. **Problem statement:** We have a solid cube with side length 10 cm. A pyramid-shaped hole is drilled inside it. The pyramid has a square base of side 5 cm and height 5 cm. We need to find the surface area and volume of the remaining solid after the hole is drilled. 2. **Formulas and important rules:** - Volume of a cube: $$V_{cube} = s^3$$ where $s$ is the side length. - Surface area of a cube: $$A_{cube} = 6s^2$$ - Volume of a pyramid: $$V_{pyramid} = \frac{1}{3} \times \text{base area} \times \text{height}$$ - Surface area of the remaining solid = original cube surface area minus the base area of the pyramid hole (since that part is removed) plus the lateral surface area of the pyramid hole (since the pyramid's inside faces become part of the solid's surface). 3. **Calculate the volume of the cube:** $$V_{cube} = 10^3 = 1000 \text{ cm}^3$$ 4. **Calculate the volume of the pyramid hole:** Base area of pyramid = $$5 \times 5 = 25 \text{ cm}^2$$ Height = 5 cm $$V_{pyramid} = \frac{1}{3} \times 25 \times 5 = \frac{125}{3} \approx 41.67 \text{ cm}^3$$ 5. **Calculate the volume of the remaining solid:** $$V_{solid} = V_{cube} - V_{pyramid} = 1000 - 41.67 = 958.33 \text{ cm}^3$$ 6. **Calculate the surface area of the cube:** $$A_{cube} = 6 \times 10^2 = 600 \text{ cm}^2$$ 7. **Calculate the lateral surface area of the pyramid hole:** - The pyramid has 4 triangular faces. - Each triangular face has base 5 cm. - Slant height $l$ is found using Pythagoras theorem: $$l = \sqrt{\left(\frac{5}{2}\right)^2 + 5^2} = \sqrt{6.25 + 25} = \sqrt{31.25} \approx 5.59 \text{ cm}$$ - Area of one triangular face: $$A_{triangle} = \frac{1}{2} \times 5 \times 5.59 = 13.975 \text{ cm}^2$$ - Total lateral area: $$A_{lateral} = 4 \times 13.975 = 55.9 \text{ cm}^2$$ 8. **Calculate the new surface area of the solid:** - The base of the pyramid hole (25 cm²) is removed from the cube surface. - The 4 triangular faces of the pyramid hole are added. $$A_{solid} = A_{cube} - 25 + 55.9 = 600 - 25 + 55.9 = 630.9 \text{ cm}^2$$ **Final answers:** - Volume of the solid after drilling the pyramid hole: $$958.33 \text{ cm}^3$$ - Surface area of the solid after drilling the pyramid hole: $$630.9 \text{ cm}^2$$