, 2022

Answer: $\frac{5\sqrt{39}}{2}$

Let the vertices of the pyramid be $A$, $B$, $C$, $D$, where $BD = 6$ and all other edges have length $5$. Let also $M$ be the midpoint of $BD$ and $N$ be the foot of perpendicular from $A$ to the base $BCD$. Of course we have $BM = MD = 3$ and $CM = 4$. Note that
$$AN^2 = AB^2 - BN^2 = AC^2 - CN^2 = AD^2 - DN^2.$$
As $AB = AC = AD = 5$, we have $BN = CN = DN$ and so $N$ is the circumcentre of $\triangle BCD$. The circumradius of $\triangle BCD$ is given by the extended sine formula
$$BN = \frac{CD}{2 \sin B} = \frac{5}{2 \cdot \frac{4}{5}} = \frac{25}{8}.$$
The area of $\triangle BCD$ is $\frac{1}{2} \cdot BD \cdot MC = 12$. The volume of the pyramid is thus
$$\frac{1}{3} \cdot 12 \cdot AN = 4 \cdot \sqrt{5^2 - \left(\frac{25}{8}\right)^2} = \frac{5\sqrt{39}}{2}.$$