Cube inscribed in a sphere

**Hellbent** · January 14th 2011, 05:05 PM

Hi,

A cube with volume 8 cubic centimeters is inscribed in a sphere so that each vertex of the cube touches the sphere. What is the length of the diameter, in centimeters, of the sphere?

A. $2$

B. $\sqrt{6} \approx 2.45$

C. $2.5$

D. $2\sqrt{3} \approx 3.46$

E. $4$

Made progress: Volume of cube = $\mathrm{side}^3$ . Then $\sqrt[3]{8} = 2$ . A diameter is the longest line segment that can be drawn in a circle.
Since this is a cube, the diagonal forms a rectangular prism. Correct? How do I find the length of a diagonal in a rectangular prism, thus finding the length of the diameter?

**Archie Meade** · January 14th 2011, 05:18 PM

Originally Posted by Hellbent

Hi,

A cube with volume 8 cubic centimeters is inscribed in a sphere so that each vertex of the cube touches the sphere. What is the length of the diameter, in centimeters, of the sphere?

A. $2$

B. $\sqrt{6} \approx 2.45$

C. $2.5$

D. $2\sqrt{3} \approx 3.46$

E. $4$

Made progress: Volume of cube = $\mathrm{side}^3$ . Then $\sqrt[3]{8} = 2$ . A diameter is the longest line segment that can be drawn in a circle.
Since this is a cube, the diagonal forms a rectangular prism. Correct? How do I find the length of a diagonal in a rectangular prism, thus finding the length of the diameter?

The centre of the cube coincides with the centre of the sphere,
so you are looking for the distance from one corner at the base of the cube to
the top corner of the opposite face on the other side.

This means you need to use Pythagoras' theorem twice,
first to find the length of the cube base diagonal,
then to find the distance between "opposite" corners.

You'll find D is the solution.

**Hellbent** · January 14th 2011, 05:46 PM

Originally Posted by Archie Meade

The centre of the cube coincides with the centre of the sphere,
so you are looking for the distance from one corner at the base of the cube to
the top corner of the opposite face on the other side.

This means you need to use Pythagoras' theorem twice,
first to find the length of the cube base diagonal,
then to find the distance between "opposite" corners.

You'll find D is the solution.

Huh?

Diagonal of Prism: $\sqrt{2^2 + 2^2 + 2^2}$

$\sqrt{12}$

$\sqrt{4}\sqrt{3}$

$2\sqrt{3}$

**DrSteve** · January 14th 2011, 05:54 PM

Doing the Pythagorean Theorem twice leads to the generalized Pythagorean Theorem:

$d^2=a^2+b^2+c^2$

Here a,b,c are the length, width and height of the prism, and d is the long diagonal of the prism.

**CaptainBlack** · January 14th 2011, 11:51 PM

Originally Posted by Hellbent

Hi,

A cube with volume 8 cubic centimeters is inscribed in a sphere so that each vertex of the cube touches the sphere. What is the length of the diameter, in centimeters, of the sphere?

A. $2$

B. $\sqrt{6} \approx 2.45$

C. $2.5$

D. $2\sqrt{3} \approx 3.46$

E. $4$

Made progress: Volume of cube = $\mathrm{side}^3$ . Then $\sqrt[3]{8} = 2$ . A diameter is the longest line segment that can be drawn in a circle.
Since this is a cube, the diagonal forms a rectangular prism. Correct? How do I find the length of a diagonal in a rectangular prism, thus finding the length of the diameter?

The diagonal of the cube is a diameter of the sphere.

CB

**Archie Meade** · January 15th 2011, 02:32 AM

Originally Posted by Hellbent

Huh?

Diagonal of Prism: $\sqrt{2^2 + 2^2 + 2^2}$

$\sqrt{12}$

$\sqrt{4}\sqrt{3}$

$2\sqrt{3}$

Exactly.

The length diagonal of the base of the cube is $\sqrt{2^2+2^2}$

and it's square is $2^2+2^2$

The distance from one corner to another through the centre is $\sqrt{\left(2^2+2^2\right)+2^2}=\sqrt{3\left(2^2\r ight)}=2\sqrt{3}$

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