1. **Problem statement:** We have a frustum formed by removing a smaller cone from a larger cone. The smaller cone has height $10$ m and top radius $6$ m. The original large cone has height $50$ m. We need to find: a) The radius of the original large cone. b) The volume of the frustum to the nearest integer. 2. **Formula for volume of a cone:** $$V = \frac{1}{3} \pi r^2 h$$ where $r$ is the radius and $h$ is the height. 3. **Step a) Calculate the radius of the original large cone:** Since the cones are similar, the ratio of corresponding dimensions is the same. Let $R$ be the radius of the original large cone. The smaller cone has radius $6$ m and height $10$ m. The large cone has height $50$ m. By similarity: $$\frac{R}{50} = \frac{6}{10}$$ Solve for $R$: $$R = 50 \times \frac{6}{10} = 50 \times 0.6 = 30$$ So, the radius of the original large cone is $30$ m. 4. **Step b) Calculate the volume of the frustum:** The frustum volume is the volume of the large cone minus the volume of the small cone. - Volume of large cone: $$V_{large} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (30)^2 (50) = \frac{1}{3} \pi \times 900 \times 50 = 15000 \pi$$ - Volume of small cone: $$V_{small} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 (10) = \frac{1}{3} \pi \times 36 \times 10 = 120 \pi$$ - Volume of frustum: $$V_{frustum} = V_{large} - V_{small} = 15000 \pi - 120 \pi = 14880 \pi$$ Calculate numerical value: $$V_{frustum} \approx 14880 \times 3.1416 = 46756.8$$ Rounded to the nearest integer: $$46757$$ **Final answers:** - Radius of original large cone: $30$ m - Volume of frustum: $46757$ cubic meters