1. **Problem statement:** We have a solid shape made up of a cone and a hemisphere. The radius of both the hemisphere and the cone is $x$ cm, and the perpendicular height of the cone is $2x$ cm. 2. **Formula for volume:** - Volume of a cone: $$V_{cone} = \frac{1}{3} \pi r^2 h$$ - Volume of a hemisphere: $$V_{hemisphere} = \frac{2}{3} \pi r^3$$ Since $r = x$ and $h = 2x$, substitute these into the formulas: $$V_{cone} = \frac{1}{3} \pi x^2 (2x) = \frac{2}{3} \pi x^3$$ $$V_{hemisphere} = \frac{2}{3} \pi x^3$$ 3. **Total volume $V$:** $$V = V_{cone} + V_{hemisphere} = \frac{2}{3} \pi x^3 + \frac{2}{3} \pi x^3 = \frac{4}{3} \pi x^3$$ 4. **Rearranging to make $x$ the subject:** Start with: $$V = \frac{4}{3} \pi x^3$$ Multiply both sides by $\frac{3}{4\pi}$: $$\frac{3V}{4\pi} = x^3$$ Take the cube root: $$x = \sqrt[3]{\frac{3V}{4\pi}}$$ 5. **Find $x$ when $V = 500$ cm³:** Substitute $V = 500$: $$x = \sqrt[3]{\frac{3 \times 500}{4 \pi}} = \sqrt[3]{\frac{1500}{4 \pi}}$$ Calculate the inside value: $$\frac{1500}{4 \pi} \approx \frac{1500}{12.566} \approx 119.366$$ Now cube root: $$x \approx \sqrt[3]{119.366} \approx 4.93 \text{ cm}$$ Convert to mm: $$4.93 \text{ cm} = 49.3 \text{ mm}$$ Rounded to the nearest mm: $$x = 49 \text{ mm}$$