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प्रश्न
A solid toy s in the form of a hemisphere surrounded by a right circular cone . The height of cone is 4 cm and the diameter of the base is 8 cm . Determine the volume of the toy. If a cube circumscribes the toy , then find the difference of the volumes of cube and the toy .
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उत्तर
The height of the cone, h = 4 cm
Diameter of the base, d = 8 cm
Radius of the cone, r = 4 cm
lateral side will be
\[l = \sqrt{h^2 + r^2}\]
\[l = \sqrt{\left( 4 \right)^2 + \left( 4 \right)^2} = 4\sqrt{2}\]
Volume of the toy = volume of hemisphere + volume of cone
\[\Rightarrow V = \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h\]
\[ = \frac{\pi r^2}{3}\left[ 2r + h \right]\]
\[ = \frac{\pi \left( 4 \right)^2}{3}\left[ 2 \times 4 + 4 \right]\]
\[ = 201 . 14 {cm}^3\]
When the cube circumscribes the toy, then
Volume of the cube = \[a^3 = \left( 8 \right)^3 = 512 c m^3\]
\[\text { Volume of cube - volume of toy = } 512 - 201 . 14 = 310 . 86 c m^3\]
Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cone
\[= 2\pi r^2 + \pi rl\]
\[ = 2\pi \left( 4 \right)^2 + \pi \times 4 \times 4\sqrt{2}\]
\[ = \pi \left( 4 \right)^2 \left[ 2 + \sqrt{2} \right]\]
\[ = 171 . 68 {cm}^2\]
