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?
Show Hint
- Rs. 3,60,000
- Rs. 3,72,000
- Rs. 1,12,000
- Rs. 5,23,000
- Rs. 4,73,000
The Correct Option is B
Approach Solution - 1
To determine the cost of gold-plating the trophy, we need to calculate the total surface area that will be plated, which includes the lateral surface area of the frustum and the surface area of the hemisphere.
- Calculate the lateral surface area of the frustum of the cone:
The formula for the lateral surface area of a frustum of a cone is given by:
\(A_{\text{frustum}} = \pi (R + r) l\)
Where \(R\) is the bottom radius, \(r\) is the top radius, and \(l\) is the slant height. The slant height can be calculated by the Pythagorean theorem:
\(l = \sqrt{h^2 + (R - r)^2}\)
Given:
- \(R = 30 \, \text{cm}\),
- \(r = 20 \, \text{cm}\),
- \(h = 40 \, \text{cm}\).
First calculate the slant height, \(l\):
\(l = \sqrt{40^2 + (30 - 20)^2} = \sqrt{1600 + 100} = \sqrt{1700} = 10\sqrt{17} \, \text{cm}\)
Thus, the lateral surface area, \(A_{\text{frustum}}\), is:
\(A_{\text{frustum}} = \pi (30 + 20) \times 10\sqrt{17} = 50\pi \times 10\sqrt{17} = 500\pi \sqrt{17} \, \text{cm}^2\)
- Calculate the surface area of the hemisphere:
The surface area of a hemisphere is given by:
\(A_{\text{hemisphere}} = 2\pi r^2\)
Given that the radius \(r\) of the hemisphere is 20 cm:
\(A_{\text{hemisphere}} = 2\pi \times 20^2 = 2\pi \times 400 = 800\pi \, \text{cm}^2\)
- Calculate the total surface area to be gold-plated:
Since the top surface of the frustum is not exposed due to the attachment of the hemisphere, the total gold-plated surface area, \(A_{\text{total}}\), is:
\(A_{\text{total}} = A_{\text{frustum}} + A_{\text{hemisphere}} = 500\pi \sqrt{17} + 800\pi\)
- Calculate the cost of gold-plating:
The cost of gold-plating per square cm is Rs. 40.
Therefore, the total cost, \(C\), is:
\(C = 40 \times (A_{\text{total}})\)
Using the approximation \(\pi \approx 3.14\) and \(\sqrt{17} \approx 4.12\):
\(C = 40 \times (500 \times 3.14 \times 4.12 + 800 \times 3.14)\)
Calculating separately:
\(500 \times 3.14 \times 4.12 = 6462.8\)
\(800 \times 3.14 = 2512\)
Total:
\(C = 40 \times (6462.8 + 2512) = 40 \times 8974.8 = 358992\)
Round this to the nearest option: Rs. 3,72,000.
Therefore, the cost for gold-plating the trophy is approximately Rs. 3,72,000.
Approach Solution -2
The surface area of the frustum and hemisphere are calculated using their respective formulae.
Step 2: Add the surface areas and multiply by the cost per square cm.
The total surface area is multiplied by the cost of Rs. 40 per square cm.
Final Answer: \[ \boxed{\text{Rs. 3,72,000}} \]
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