+194 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.)
+9466 Let a be the length of an edge of the tetrahedron.
A face of the cube looks like this:
By the Pythagorean theorem,
62 + 62 = a2
Combine like terms.... n + n = n * 2 so 62 + 62 = 62 * 2
62 * 2 = a2
Take the positive square root of both sides of the equation.
√[ 62 * 2 ] = a
√62 * √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
+26367 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
