Frustum Cone F2Ed61

1. **State the problem:** We have a frustum ABQP formed by removing a smaller cone VAB from a larger cone VPQ. The radius of the smaller cone is $5$ cm, the radius of the larger cone is $6$ cm, and the height of the frustum is $4$ cm. Triangle $VXB$ is similar to triangle $VYQ$. We need to find: (a) The height of the smaller cone $VAB$. (b) The curved surface area of the frustum. 2. **Use similarity of triangles:** Since triangles $VXB$ and $VYQ$ are similar, the ratios of corresponding sides are equal. Let the height of the smaller cone be $h$ cm. The height of the larger cone is $h + 4$ cm (since the frustum height is $4$ cm). The radii correspond to the bases of the cones: smaller cone radius $= 5$ cm, larger cone radius $= 6$ cm. By similarity: $$\frac{h}{h+4} = \frac{5}{6}$$ 3. **Solve for $h$:** Multiply both sides by $h+4$: $$h = \frac{5}{6}(h+4)$$ Expand right side: $$h = \frac{5}{6}h + \frac{20}{6}$$ Bring terms involving $h$ to one side: $$h - \frac{5}{6}h = \frac{20}{6}$$ Simplify left side: $$\frac{6}{6}h - \frac{5}{6}h = \frac{20}{6}$$ $$\frac{1}{6}h = \frac{20}{6}$$ Multiply both sides by $6$: $$h = 20$$ So, the height of the smaller cone is $20$ cm. 4. **Find the height of the larger cone:** $$h + 4 = 20 + 4 = 24 \text{ cm}$$ 5. **Calculate the slant heights:** Slant height $l$ of a cone is given by: $$l = \sqrt{r^2 + h^2}$$ For the larger cone: $$l_{large} = \sqrt{6^2 + 24^2} = \sqrt{36 + 576} = \sqrt{612} = 6\sqrt{17}$$ For the smaller cone: $$l_{small} = \sqrt{5^2 + 20^2} = \sqrt{25 + 400} = \sqrt{425} = 5\sqrt{17}$$ 6. **Calculate the curved surface area (CSA) of the frustum:** CSA of frustum = CSA of larger cone $-$ CSA of smaller cone CSA of cone = $\pi r l$ So, $$\text{CSA}_{frustum} = \pi \times 6 \times 6\sqrt{17} - \pi \times 5 \times 5\sqrt{17}$$ $$= 36\pi \sqrt{17} - 25\pi \sqrt{17} = 11\pi \sqrt{17}$$ 7. **Final answers:** (a) Height of smaller cone $VAB$ is $20$ cm. (b) Curved surface area of the frustum is $11\pi \sqrt{17}$ cm$^2$. Numerical approximation: $$\sqrt{17} \approx 4.1231$$ $$11 \times \pi \times 4.1231 \approx 11 \times 3.1416 \times 4.1231 \approx 142.5$$ So, CSA $\approx 142.5$ cm$^2$.