Two solid hemispheres are joined together to make a sphere. The total surface area of solids reduced by approximately ____%.
33.33
The total surface area of one solid hemisphere of radius \(r\) is \(3\pi r^2\) (curved surface area \(2\pi r^2\) plus flat circular base \(\pi r^2\)).
For two hemispheres, the combined total surface area before joining is \(2 \times 3\pi r^2 = 6\pi r^2\).
When joined, the two flat faces (each of area \(\pi r^2\)) are glued together and become internal — they no longer form part of the surface. The resulting sphere has surface area \(4\pi r^2\).
Reduction in surface area \(= 6\pi r^2 - 4\pi r^2 = 2\pi r^2\).
Percentage reduction \(= \frac{2\pi r^2}{6\pi r^2} \times 100 = \frac{1}{3} \times 100 \approx 33.33\%\).
Hence, the total surface area is reduced by approximately 33.33%.
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Important Questions from 3-D Mensuration
Find the total surface area of a closed cylinder having a base radius of 70 m and a height of 110 m. [Use π = \(22\over7\)]
- The curved surface area of a cylinder is half of its total surface area. If its height is 195 cm, then find its diameter (in cm).
The diameter of the base and slant height of a right circular cone are 30 cm and 113 cm, respectively. Find the volume (in cm³) of the given cone.
(Use $\pi = \frac{22}{7}$)- A cylindrical rod has an curved surface area of $4,900 \text{ cm}^2$. If the length of the rod is 97 cm, then the radius (in cm) of the rod, correct to two places of decimal, is:
Take $\pi = \frac{22}{7}$ - The volume (in $m^3$) of a cube, each of whose edges is 42 m, is:
