ABC is a triangle right-angled at B with AB=8\text{ cm} and BC=6\text{ cm}. It is made to revolve about its side BC. What is the approximate total surface area of the cone so formed? (take \pi=\frac{22}{7})
- A. 452\text{ cm}^{2} ✓
- B. 440\text{ cm}^{2}
- C. 432\text{ cm}^{2}
- D. 420\text{ cm}^{2}
Correct Answer: A. 452\text{ cm}^{2}
Explanation
Revolving around BC makes BC the height (h=6) and AB the radius (r=8). Slant height l = \sqrt{8^2 + 6^2} = 10. Total surface area = \pi r(l+r) = \frac{22}{7} \times 8 \times (10+8) = \frac{22}{7} \times 144 \approx 452.57\text{ cm}^2, which approximates to 452\text{ cm}^2.
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