The internal and external radii of a hollow hemisphere are \(5\sqrt{2} \text{ cm}\) and \(10 \text{ cm}\) respectively. A cone of height \(5\sqrt{7} \text{ cm}\) and radius \(5\sqrt{2} \text{ cm}\) is surmounted on the hemisphere as shown in the figure. Find the total surface area of the object in terms of \(\pi\). (Use \(\sqrt{2} = 1.4\))
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For hollow objects, "total surface area" includes both internal and external visible surfaces unless it's explicitly described as a closed solid. Don't forget the flat ring connecting the inner and outer shells.
Step 1: Understanding the Concept:
The total surface area (TSA) of the composite object is the sum of all exposed surfaces:
1. The curved surface area of the cone.
2. The area of the top ring (between external and internal radii).
3. The external curved surface area of the hemisphere.
4. The internal curved surface area of the hemisphere. Step 2: Key Formula or Approach:
- Slant height of cone (\(l\)) = \(\sqrt{h^2 + r^2}\)
- CSA of cone = \(\pi r l\)
- Area of ring = \(\pi(R^2 - r^2)\)
- CSA of hemisphere = \(2 \pi r^2\) Step 3: Detailed Explanation:
Given: External radius \(R = 10 \text{ cm}\), Internal radius \(r = 5\sqrt{2} \text{ cm}\), Cone height \(h = 5\sqrt{7} \text{ cm}\), Cone radius = \(5\sqrt{2} \text{ cm}\).
1. Slant height of cone:
\[ l = \sqrt{(5\sqrt{7})^2 + (5\sqrt{2})^2} = \sqrt{175 + 50} = \sqrt{225} = 15 \text{ cm} \]
2. Exposed Surfaces:
- CSA of cone: \(\pi \times 5\sqrt{2} \times 15 = 75\sqrt{2} \pi \text{ cm}^2\).
Using \(\sqrt{2} = 1.4\): \(75 \times 1.4 \pi = 105 \pi \text{ cm}^2\).
- Area of flat ring: \(\pi(10^2 - (5\sqrt{2})^2) = \pi(100 - 50) = 50 \pi \text{ cm}^2\).
- Outer CSA of hemisphere: \(2 \pi R^2 = 2 \pi (100) = 200 \pi \text{ cm}^2\).
- Inner CSA of hemisphere: \(2 \pi r^2 = 2 \pi (50) = 100 \pi \text{ cm}^2\).
Total Surface Area = \(105\pi + 50\pi + 200\pi + 100\pi = 455 \pi \text{ cm}^2\). Step 4: Final Answer:
The total surface area of the object is \(455 \pi \text{ cm}^2\).
