A cubical block of side 14 cm is surmounted by a hemisphere of radius 7 cm. What is the total surface area of the solid thus formed? (take \pi=\frac{22}{7})
- A. 1330\text{ cm}^{2} ✓
- B. 1306\text{ cm}^{2}
- C. 1296\text{ cm}^{2}
- D. 1256\text{ cm}^{2}
Correct Answer: A. 1330\text{ cm}^{2}
Explanation
Total surface area = (Surface area of cube) + (Curved Surface Area of hemisphere) - (Area of base of hemisphere). Total area = 6a^2 + 2\pi r^2 - \pi r^2 = 6a^2 + \pi r^2. Substituting a=14 and r=7: 6(14)^2 + \frac{22}{7}(7)^2 = 1176 + 154 = 1330\text{ cm}^2.
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