Question

A solid trophy, consisting of two parts, has been designed in the following manner: the bottom part is a frustum of a cone with the bottom radius 30 cm, the top radius 20 cm, and height 40 cm, while the top part is a hemisphere with radius 20 cm. Moreover, the flat surface of the hemisphere is the same as the top surface of the frustum.

If the entire trophy is to be gold-plated at the cost of Rs. 40 per square cm, what would the cost for gold-plating be closest to?

Accepted Answer

C. The surface area to be gold-plated consists of: 1) Curved surface area of frustum = $\pi (r_1 + r_2) l$, where $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1700} = 10\sqrt{17} \approx 41.23$ cm. $	ext{CSA}_{frustum} = \pi (30 + 20) (10\sqrt{17}) = 500\pi \sqrt{17} \approx 500 	imes 3.1416 	imes 41.231 = 64766.3$ sq cm. 2) Base area of frustum = $\pi r_1^2 = \pi (30)^2 = 900\pi \approx 2827.4$ sq cm. 3) Curved surface area of hemisphere = $2\pi r_2^2 = 2\pi (20)^2 = 800\pi \approx 2513.3$ sq cm. Total Surface Area = $500\pi \sqrt{17} + 1700\pi = \pi (500\sqrt{17} + 1700) \approx 3.1416 	imes (2061.55 + 1700) = 3.1416 	imes 3761.55 \approx 11817.3$ sq cm. Total Cost = $11817.3 	imes 40 \approx 472,690 \approx 	ext{Rs. } 4,73,000$.

Cost of Gold Plating a Solid Trophy

Explanation

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