1. **Problem statement:** A cone is divided into a small cone and a frustum. The curved surface area of the small cone is $15\pi$ cm², the curved surface area of the frustum is $120\pi$ cm², and the volume of the small cone is $12\pi$ cm³. We need to find the volume of the frustum. 2. **Formulas and important rules:** - Curved surface area (CSA) of a cone: $$\text{CSA} = \pi r l$$ where $r$ is the radius of the base and $l$ is the slant height. - Volume of a cone: $$V = \frac{1}{3} \pi r^2 h$$ where $h$ is the height. - The frustum is formed by cutting the small cone from the large cone, so the total curved surface area of the large cone is the sum of the small cone and frustum curved surface areas. 3. **Assign variables:** Let the large cone have radius $R$, slant height $L$, and height $H$. Let the small cone have radius $r$, slant height $l$, and height $h$. 4. **Given data:** - Small cone curved surface area: $$\pi r l = 15\pi \implies r l = 15$$ - Frustum curved surface area: $$\pi (R + r)(L - l) = 120\pi \implies (R + r)(L - l) = 120$$ - Volume of small cone: $$\frac{1}{3} \pi r^2 h = 12\pi \implies r^2 h = 36$$ 5. **Relate dimensions:** Since the small cone is similar to the large cone (both cones), the ratios of corresponding dimensions are equal: $$\frac{r}{R} = \frac{l}{L} = \frac{h}{H} = k$$ for some scale factor $k$. 6. **Express small cone dimensions in terms of large cone:** $$r = k R, \quad l = k L, \quad h = k H$$ 7. **Use the small cone curved surface area:** $$r l = 15 \implies (k R)(k L) = k^2 R L = 15$$ 8. **Use the frustum curved surface area:** The frustum curved surface area is: $$\pi (R + r)(L - l) = 120\pi \implies (R + r)(L - l) = 120$$ Substitute $r = k R$ and $l = k L$: $$(R + k R)(L - k L) = R(1 + k) L(1 - k) = R L (1 + k)(1 - k) = R L (1 - k^2) = 120$$ 9. **From steps 7 and 8, we have:** $$k^2 R L = 15$$ $$R L (1 - k^2) = 120$$ Add these two equations: $$k^2 R L + R L (1 - k^2) = 15 + 120$$ $$R L (k^2 + 1 - k^2) = 135$$ $$R L = 135$$ 10. **Find $k^2$:** From step 7: $$k^2 R L = 15 \implies k^2 \times 135 = 15 \implies k^2 = \frac{15}{135} = \frac{1}{9}$$ 11. **Find $k$:** $$k = \frac{1}{3}$$ 12. **Find volume of large cone:** Recall volume of small cone: $$r^2 h = 36$$ Substitute $r = k R$, $h = k H$: $$(k R)^2 (k H) = k^3 R^2 H = 36$$ Since $k = \frac{1}{3}$: $$\left(\frac{1}{3}\right)^3 R^2 H = 36 \implies \frac{1}{27} R^2 H = 36 \implies R^2 H = 972$$ 13. **Volume of large cone:** $$V_{large} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi \times 972 = 324\pi$$ 14. **Volume of frustum:** $$V_{frustum} = V_{large} - V_{small} = 324\pi - 12\pi = 312\pi$$ **Final answer:** $$\boxed{312\pi \text{ cm}^3}$$