What is the whole surface area of the frustum?
Consider the following for the next two (02) items that follow :<br>A frustum of a right cone has a top of diameter 2k, bottom of diameter 2.5k and height k.
- A. 39\pi k^{2}/8 ✓
- B. 41\pi k^{2}/8
- C. 43\pi k^{2}/8
- D. 45\pi k^{2}/8
Correct Answer: A. 39\pi k^{2}/8
Explanation
Top radius r=k, bottom radius R=1.25k. Slant height l = \sqrt{h^2 + (R-r)^2} = \sqrt{k^2 + (0.25k)^2} = \frac{\sqrt{17}}{4}k. Total Surface Area = \pi r^2 + \pi R^2 + \pi(R+r)l = \pi k^2 (1 + \frac{25}{16} + \frac{9\sqrt{17}}{16}) = \frac{\pi k^2}{16}(41 + 9\sqrt{17}). Using \sqrt{17} \approx 4.12, TSA \approx 4.88 \pi k^2, which is closest to \frac{39}{8}\pi k^2 = 4.875 \pi k^2.
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