1. **State the problem:** We have a large cone with height 15 cm and base radius 6 cm. Inside it, a smaller similar cone with base radius $x$ cm is removed, creating a frustum. The volume of the frustum is given as $\frac{4212}{25}\pi$ cm³. We need to find $x$. 2. **Recall volume formula for a cone:** $$V = \frac{1}{3}\pi r^2 h$$ 3. **Set variables:** - Large cone radius $R = 6$ cm - Large cone height $H = 15$ cm - Small cone radius $r = x$ cm - Small cone height $h$ (unknown) 4. **Use similarity of cones:** Since the small cone is similar to the large cone, $$\frac{r}{R} = \frac{h}{H} \implies \frac{x}{6} = \frac{h}{15} \implies h = \frac{15x}{6} = \frac{5x}{2}$$ 5. **Calculate volumes:** - Volume of large cone: $$V_{large} = \frac{1}{3}\pi (6)^2 (15) = \frac{1}{3}\pi \times 36 \times 15 = 180\pi$$ - Volume of small cone: $$V_{small} = \frac{1}{3}\pi x^2 h = \frac{1}{3}\pi x^2 \times \frac{5x}{2} = \frac{5}{6}\pi x^3$$ 6. **Volume of frustum:** $$V_{frustum} = V_{large} - V_{small} = 180\pi - \frac{5}{6}\pi x^3$$ Given: $$V_{frustum} = \frac{4212}{25}\pi$$ 7. **Set up equation and solve for $x$:** $$180\pi - \frac{5}{6}\pi x^3 = \frac{4212}{25}\pi$$ Divide both sides by $\pi$: $$180 - \frac{5}{6} x^3 = \frac{4212}{25}$$ Rearranged: $$180 - \frac{4212}{25} = \frac{5}{6} x^3$$ Calculate left side: $$180 = \frac{180 \times 25}{25} = \frac{4500}{25}$$ $$\frac{4500}{25} - \frac{4212}{25} = \frac{288}{25}$$ So: $$\frac{288}{25} = \frac{5}{6} x^3$$ Multiply both sides by $\frac{6}{5}$: $$x^3 = \frac{288}{25} \times \frac{6}{5} = \frac{288 \times 6}{25 \times 5} = \frac{1728}{125}$$ 8. **Find cube root:** $$x = \sqrt[3]{\frac{1728}{125}} = \frac{\sqrt[3]{1728}}{\sqrt[3]{125}} = \frac{12}{5} = 2.4$$ **Final answer:** $$x = 2.4 \text{ cm}$$