Highly Irregular

If we had a square of side $5$ , the diagonal would be $7.07107..$ . By making this diagonal exactly $7$ , we form a rhombus, which we can crease along the shorter diagonal. Putting two such rhombuses together makes for a nearly flat tetrahedron. A little effort gets us the volume

$\dfrac {49 \sqrt{2}}{12}$

as compared with the volume of a regular tetrahedron of edge $5$

$\dfrac {125 \sqrt{2}}{12}$

Edit: Here's the work to get us the volume

$\sqrt { { 5 }^{ 2 }-2{ \left( \dfrac { 7 }{ 2 } \right) }^{ 2 } }\displaystyle \int _{ 0 }^{ 1 }{ 7h\left( 7-7h \right) } dh=\dfrac { 49\sqrt { 2 } }{ 12 }$

where the limits of the integration is based on using altitude $h$ as a parameter arbitrarily from $0$ to $1$ , the area being a rectangle varying from $0 \times 7$ to $\frac {7}{2} \times \frac {7}{2}$ to $7 \times 0$ , while the coefficient is the ratio of the actual distance between the edges of length $7$ to the parameter of arbitrary length $1$ , thus scaling the volume.

Edit 2: Here's a graphic of this squeezed tetrahedron. The top and bottom of this are the long edges of length $7$ , all the rest having length $5$ . The distance between the long edges is $\dfrac{1}{\sqrt{2}}$ , so this tetrahedron is nearly flat. You can see how two folded rhombuses can be put together to form this tetrahedron.

Note: Finding the volume of this particular tetrahedron is easy because of its symmetry. But, in the general case, given the six arbitrary edge lengths of a tetrahedron, $a, b, c, d, e, f$ , finding the volume is usually a gargantuan task. I suspect that Enlow did not go that way.

One way to find the volume is using analytical geometry with base triangle on XY plane, z coordinate of the fourth vertex being the height. There is also a formula for the volume of tetrahedron.

$V=\frac{1}{3!}\sqrt{\frac{1}{2^3} det {a_{ij}^2, 1 \choose 1, 0}}$

$a_{ij}$ denotes an edge connecting vertex $A_{i}$ and $A_{j}$ and

$det {a_{ij}^2, 1 \choose 1, 0} = det \begin{pmatrix} 0 & a_{1,2}^2 & a_{1,3}^2 & a_{1,4}^2 & 1 \\ a_{1,2}^2 & 0 & a_{2,3}^2 & a_{2,4}^2 & 1 \\ a_{1,3}^2 & a_{2,3}^2 & 0 & a_{3,4}^2 & 1 \\ a_{1,4}^2 & a_{2,4}^2 & a_{3,4}^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}$

In our case there are 11 possible configurations:

No of Length 7 Edges | No of Configurations | Volumes

               0          1                               14.73 

               1          1                               14.87 

               2          2                               18.04 5.77    

               3          3                               23.01 20.82   18.26

               4          2                               25.25 25.17   

               5          1                               32.22 

               6          1                               40.42

The configuration with 2 long edges opposite each other without common vertex gives lowest determinant:

$det \begin{pmatrix} 0 & 5^2 & 5^2 & 7^2 & 1 \\ 5^2 & 0 & 7^2 & 5^2 & 1 \\ 5^2 & 7^2 & 0 & 5^2 & 1 \\ 7^2 & 5^2 & 5^2 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix} = 9604$

resulting with volume value of $\frac{49 \sqrt{2}}{12} \approx 5.77$ .