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. Total surface area to be plated = Curved surface area of frustum + Curved surface area of hemisphere + Bottom base area of frustum. 1. Slant height of frustum $l = \sqrt{h^2 + (r_1 - r_2)^2} = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1600 + 100} = \sqrt{1700} = 10\sqrt{17} \approx 41.231$ cm. 2. CSA of frustum = $\pi (r_1 + r_2) l = \pi (30 + 20) (10\sqrt{17}) = 500\sqrt{17}\pi \approx 500 	imes 41.231 	imes 3.1416 \approx 6477$ cm$^2$. 3. CSA of hemisphere = $2\pi r^2 = 2 \pi (20^2) = 800\pi \approx 2513.27$ cm$^2$. 4. Base area = $\pi R^2 = \pi (30^2) = 900\pi \approx 2827.43$ cm$^2$. Total Area = $\pi (500\sqrt{17} + 800 + 900) = \pi (2061.55 + 1700) = 3761.55 \pi \approx 11817.2$ cm$^2$. Cost = $11817.2 	imes 40 \approx 	ext{Rs. } 4,72,688 \approx 	ext{Rs. } 4,73,000$.

Trophy Gold-Plating Cost

Explanation

Practise this under exam conditions