Question

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 .

Answer 1

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

\[= 2\pi r^2 + \pi rl\]

\[ = 2\pi \left( 4 \right)^2 + \pi \times 4 \times 4\sqrt{2}\]

\[ = \pi \left( 4 \right)^2 \left[ 2 + \sqrt{2} \right]\]

\[ = 171 . 68 {cm}^2\]