1. **State the problem:** Find the volume of the composite solid consisting of a cone on top of a frustum-like solid. The cone has height $8$ in, the frustum has height $10$ in, the bottom radius of the frustum is $9$ in, and the top radius of the frustum (which is also the base radius of the cone) is $4$ in. 2. **Formulas used:** - Volume of a cone: $$V_{cone} = \frac{1}{3} \pi r^2 h$$ - Volume of a frustum of a cone: $$V_{frustum} = \frac{1}{3} \pi h (r_1^2 + r_1 r_2 + r_2^2)$$ where $r_1$ and $r_2$ are the radii of the two circular faces. 3. **Calculate the volume of the cone:** - Radius $r = 4$ in, height $h = 8$ in - $$V_{cone} = \frac{1}{3} \pi (4)^2 (8) = \frac{1}{3} \pi \times 16 \times 8 = \frac{128}{3} \pi$$ cubic inches 4. **Calculate the volume of the frustum:** - Bottom radius $r_1 = 9$ in, top radius $r_2 = 4$ in, height $h = 10$ in - $$V_{frustum} = \frac{1}{3} \pi (10) (9^2 + 9 \times 4 + 4^2) = \frac{10}{3} \pi (81 + 36 + 16) = \frac{10}{3} \pi (133) = \frac{1330}{3} \pi$$ cubic inches 5. **Calculate total volume:** - $$V_{total} = V_{cone} + V_{frustum} = \frac{128}{3} \pi + \frac{1330}{3} \pi = \frac{1458}{3} \pi = 486 \pi$$ cubic inches 6. **Final answer:** - $$\boxed{486 \pi \text{ cubic inches}}$$ which is approximately $1527.88$ cubic inches.