# Calculating volume of a spaceA spherical holiday ornament is to be packaged in a cubical box so that all sides touch the sphere.if the volume of the sphere is 288(pie)cm^3,find the amount of left...

A spherical holiday ornament is to be packaged in a cubical box so that all sides touch the sphere.if the volume of the sphere is 288(pie)cm^3,find the amount of left over space in the cubical box to the nearest tenth of a centimeter

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Another interesting question.

If four chocolate spheres with a volumes of 4*pi/3 are inside a tetrahedron in a way that each of the four spheres touch the three other spheres; and each of the sides (is the correct word faces?) touch three of the spheres, what is the volume of the tetrahedron, and what is the volume of the space inside the tetrahedron, but outside the spheres?

I just love chocolate, don't you? I'd rather be in chocolate than in love.

This is an interesting question rather than a discussion topic. The method for finding the volume of the left over space between the sphere and the cube is equal to the difference in their volumes. If the sphere fits into the cube than the diameter of the sphere should be equal to the side of the cube.

The volume of sphere is given to be 288.pi cm^3 and if radius of sphere is 'r' than it is equal to (4/3).pi.r^3. This gives 'r' equal to 6cm. Therefore the diameter of the sphere is 12cm. This sphere will fit into a cube of side 12cm. The volume of cube having side 12cm is 12^3 or 1728cm^3.

The volume of left over space between cube and sphere is equal to 1728 minus 288.pi cm^3 or 823.2 cm^3.

The post 1 by Pee2 states that the left over space is to be determined to the nearest tenth of a centimeter. Here the poster has made a typo mistake because volume or space is measured in cubic units so it has to be to the nearest tenth of the cubic centimeter or cm^3.

thanks for your help sir.

This is an interesting question rather than a discussion topic. The method for finding the volume of the left over space between the sphere and the cube is equal to the difference in their volumes. If the sphere fits into the cube than the diameter of the sphere should be equal to the side of the cube.

The volume of sphere is given to be 288.pi cm^3 and if radius of sphere is 'r' than it is equal to (4/3).pi.r^3. This gives 'r' equal to 6cm. Therefore the diameter of the sphere is 12cm. This sphere will fit into a cube of side 12cm. The volume of cube having side 12cm is 12^3 or 1728cm^3.

The volume of left over space between cube and sphere is equal to 1728 minus 288.pi cm^3 or 823.2 cm^3.

The post 1 by Pee2 states that the left over space is to be determined to the nearest tenth of a centimeter. Here the poster has made a typo mistake because volume or space is measured in cubic units so it has to be to the nearest tenth of the cubic centimeter or cm^3.