A wooden cubical die is formed by forming hemispherical depressions on each face of the cube such that face 1 has one depression, face 2 has two depressions and so on. The sum of number of hemispherical depressions on opposite faces is always 7. If the edge of the cubical die measures 5 cm and each hemispherical depression is of diameter 1.4 cm, find the total surface area of the die so formed.
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Given:
- Edge of cube = 5 cm
- Diameter of each hemispherical depression = 1.4 cm, so radius r = 0.7 cm
- Opposite faces sum to 7: pairs are (1,6), (2,5), (3,4) → total depressions = 1+2+3+4+5+6 = 21
Total Surface Area of the die:
TSA = TSA of cube − (area of 21 circular bases) + (CSA of 21 hemispheres)
Each circular base area = πr² and CSA of one hemisphere = 2πr², so each depression contributes: 2πr² − πr² = πr² (net addition per depression).
TSA of cube = 6 × (5)² = 150 cm²
Net area added per depression = πr² = $\dfrac{22}{7} × 0.7 × 0.7$ = 1.54 cm²
Total change = 21 × 1.54 = 32.34 cm²
$$\text{Total Surface Area} = 150 + 32.34 = \boxed{182.34 \text{ cm}^2}$$
Source: Surface Area of a Combination of Solids, Chapter 12
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Explanation
- The key idea (from Chapter 12) is: when a depression is made, you subtract the flat circular base (πr²) but add the curved surface of the hemisphere (2πr²). Net gain per depression = πr².
- Opposite faces summing to 7 is just standard die convention — use it only to confirm total = 21 (sum 1 to 6).
- Don't forget to count ALL 21 depressions across all 6 faces.
- Examiners award marks for: correct formula setup, correct r value, correct count (21), and final answer.