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A cube has side 6 lengths. Its vertices are alternately colored black and purple, as shown below. What is the volume of the tetrahedron whose corners are the purple vertices of the cube? (A tetrahedron is a pyramid with a triangular base.)

Confusedperson Jun 21, 2018

#1**+1 **

Let a be the length of an edge of the tetrahedron.

A face of the cube looks like this:

By the Pythagorean theorem,

6^{2} + 6^{2} = a^{2}

Combine like terms.... n + n = n * 2 so 6^{2} + 6^{2} = 6^{2} * 2

6^{2} * 2 = a^{2}

Take the positive square root of both sides of the equation.

√[ 6^{2} * 2 ] = a

√6^{2} * √2 = a

6√2 = a

Since each face of the cube is the same, each edge of the tetrahedron is the same.

So the tetrahedron is a regular tetrahedron.

The formula for the volume of a regular tetrahedron is

volume = (edge length)^{3} / ( 6√2 )

Plug in 6√2 for the edge length.

volume = ( 6√2 )^{3} / ( 6√2 )

Simplify.

volume = ( 6√2 )( 6√2 )( 6√2 ) / ( 6√2 )

volume = ( 6√2 )( 6√2 )

volume = 36 * 2

volume = 72 cubic units

hectictar Jun 21, 2018

#2**+1 **

**A cube has side 6 lengths. Its vertices are alternately colored black and purple, as shown below. What is the volume of the tetrahedron whose corners are the purple vertices of the cube? (A tetrahedron is a pyramid with a triangular base.)**

A cube has side 6 lengths**. \(\mathbf{\text{Let $s = 6$}}\)**

The cube has volume \(\mathbf{V_{\text{cube}} = s^3}\)

The **4** right triangular pyramids that must be carved off the cube to produce the regular tetrahedron each have volume

\(\mathbf{V_{\text{right triangular pyramid}} = \frac13\ \times \frac{s^2}{2} \times s } \)

The volume of the regular tetrahedron is \(\mathbf{ V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} }\)

\(\begin{array}{|rcll|} \hline && V_{\text{tetrahedron}} \\\\ &=& V_{\text{cube} }- 4\times V_{\text{right triangular pyramid}} \\\\ &=& s^3 - 4\times \dfrac13\ \times \dfrac{s^2}{2} \times s \\\\ &=& s^3 - \dfrac23 s^3 \\\\ &=& \dfrac13 s^3 \quad & | \quad s = 6 \\\\ &=& \dfrac{6^3}{3} \\\\ &=& \dfrac{216}{3} \\\\ &=& 72 \\ \hline \end{array}\)

The volume of the tetrahedron is **72**

heureka Jun 21, 2018