1. **State the problem:** We have a large cone with height $36$ cm and radius $10$ cm. The formula for its volume is $V = \frac{1}{3} \pi r^2 h$. A smaller cone is cut from the top parallel to the base, such that its volume is half the original cone's volume. This smaller cone is melted and formed into two spheres whose radii are in the ratio $1 : 2$. We need to find the radius of the smaller sphere. 2. **Calculate the volume of the original cone:** $$V = \frac{1}{3} \pi (10)^2 (36) = \frac{1}{3} \pi \times 100 \times 36 = 1200 \pi \text{ cm}^3$$ 3. **Volume of the smaller cone:** It is half the original volume, so: $$V_{small} = \frac{1}{2} \times 1200 \pi = 600 \pi \text{ cm}^3$$ 4. **Find the scale factor of the smaller cone:** Since the smaller cone is similar to the original (cut parallel to the base), volumes scale by the cube of the linear scale factor $k$: $$k^3 = \frac{V_{small}}{V_{original}} = \frac{600 \pi}{1200 \pi} = \frac{1}{2}$$ So, $$k = \sqrt[3]{\frac{1}{2}} = \frac{1}{\sqrt[3]{2}}$$ 5. **Find the dimensions of the smaller cone:** Height of smaller cone: $$h_{small} = k \times 36 = 36 \times \frac{1}{\sqrt[3]{2}}$$ Radius of smaller cone: $$r_{small} = k \times 10 = 10 \times \frac{1}{\sqrt[3]{2}}$$ 6. **Volume of the smaller cone is melted to make two spheres of radius ratio 1:2.** Let radius of smaller sphere be $r$, then larger sphere's radius is $2r$. Volumes of spheres: $$V_1 = \frac{4}{3} \pi r^3$$ $$V_2 = \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi 8r^3 = \frac{32}{3} \pi r^3$$ Total volume from both spheres: $$V_{total} = V_1 + V_2 = \frac{4}{3} \pi r^3 + \frac{32}{3} \pi r^3 = \frac{36}{3} \pi r^3 = 12 \pi r^3$$ 7. **Equate total sphere volume to volume of smaller cone:** $$12 \pi r^3 = 600 \pi$$ Divide both sides by $\pi$: $$12 r^3 = 600$$ $$r^3 = \frac{600}{12} = 50$$ 8. **Solution for $r$:** $$r = \sqrt[3]{50}$$ Approximate: $$r \approx 3.684 \text{ cm}$$ **Final answer:** Radius of the smaller sphere is approximately $3.68$ cm.