1. **Problem statement:** We have a frustum formed by removing a small cone from a larger similar cone. The small cone has radius $6$ m and height $10$ m. The frustum height is $50$ m. We need to find: a) The radius of the original large cone. b) The volume of the frustum to the nearest integer. 2. **Understanding similarity:** Since the small cone and the original large cone are similar, their corresponding dimensions are proportional. 3. **Calculate the radius of the original large cone:** Let the radius of the original large cone be $R$ and its height be $H$. We know the small cone height $h = 10$ m and radius $r = 6$ m. The frustum height is $50$ m, so the total height of the large cone is: $$H = h + 50 = 10 + 50 = 60$$ Using similarity of triangles: $$\frac{r}{h} = \frac{R}{H} \implies \frac{6}{10} = \frac{R}{60}$$ Cross-multiplied: $$6 \times 60 = 10 \times R$$ $$360 = 10R$$ Divide both sides by 10: $$\cancel{10}R = \frac{360}{\cancel{10}}$$ $$R = 36$$ So, the radius of the original large cone is $36$ m. 4. **Calculate the volume of the frustum:** The volume of a cone is: $$V = \frac{1}{3} \pi r^2 h$$ Volume of the large cone: $$V_{large} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (36)^2 (60)$$ Calculate: $$V_{large} = \frac{1}{3} \pi \times 1296 \times 60 = \frac{1}{3} \pi \times 77760 = 25920 \pi$$ Volume of the 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 the frustum: $$V_{frustum} = V_{large} - V_{small} = 25920 \pi - 120 \pi = 25800 \pi$$ Approximate using $\pi \approx 3.1416$: $$V_{frustum} \approx 25800 \times 3.1416 = 81036.48$$ Rounded to the nearest integer: $$\boxed{81036}$$ **Final answers:** - Radius of the original large cone: $36$ m - Volume of the frustum: $81036$ cubic meters