The internal and external radii of a hollow hemisphere are $5\sqrt{2}$ cm and 10 cm respectively. A cone of height $5\sqrt{7}$ cm and radius $5\sqrt{2}$ cm is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of $\pi$. (Use $\sqrt{2} = 1.4$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding rag
Model Answer
Given:
- Internal radius of hemisphere, r = 5√2 cm
- External radius of hemisphere, R = 10 cm
- Cone: height h = 5√7 cm, base radius = 5√2 cm (= r)
Slant height of cone:
$$l = \sqrt{r^2 + h^2} = \sqrt{(5\sqrt{2})^2 + (5\sqrt{7})^2} = \sqrt{50 + 175} = \sqrt{225} = 15 \text{ cm}$$
Total Surface Area = CSA of cone + CSA of outer hemisphere + CSA of inner hemisphere + Ring area (annular base)
- CSA of cone = $\pi r l = \pi (5\sqrt{2})(15) = 75\sqrt{2}\,\pi$
- CSA of outer hemisphere = $2\pi R^2 = 2\pi(10)^2 = 200\pi$
- CSA of inner hemisphere = $2\pi r^2 = 2\pi(5\sqrt{2})^2 = 100\pi$
- Annular ring area = $\pi(R^2 - r^2) = \pi(100 - 50) = 50\pi$
TSA $= 75\sqrt{2}\,\pi + 200\pi + 100\pi + 50\pi$
$$= 75(1.4)\pi + 350\pi = 105\pi + 350\pi = \boxed{455\pi \text{ cm}^2}$$
Source: Surface Area of a Combination of Solids, Chapter 12
---
Explanation
- The visible surfaces are: CSA of cone (top), CSA of outer hemisphere (outside), CSA of inner hemisphere (inside hollow), and the annular ring at the base (the flat circular border between inner and outer radius).
- The base of the cone coincides with the inner hollow opening, so no extra circular base is added for the cone.
- Key step: slant height l = 15 cm (clean calculation — check this first in the exam).
- Use √2 = 1.4 only at the final step: 75√2 = 75 × 1.4 = 105.
- Examiners award marks for identifying all four surface components correctly and computing l accurately.