7.1. In a triangular pyramid $A B C D$, it is known that: $A B=C D=6, A D=B C=10, \angle A B C=120^{\circ}$. Find $R^{2}$, where $R$ is the radius of the smallest sphere that can contain such a pyramid.
Answer: 49

Solution. Since the segment $AC$ fits inside the sphere, $2R \geqslant AC$. On the other hand, the sphere constructed with $AC$ as its diameter covers both the triangle $ABC$, since $\angle ABC > 90^{\circ}$, and the congruent (by three sides) triangle $BAD$, and thus covers the entire tetrahedron. Therefore, the desired value of $R$ is $AC / 2$, and $R^{2} = AC^{2} / 4$.

Next, by the cosine rule we have $AC^{2} = AB^{2} + BC^{2} - 2 AB \cdot BC \cdot \cos 120^{\circ} = 6^{2} + 10^{2} + 2 \cdot 6 \cdot 10 / 2 = 4(3^{2} + 5^{2} + 3 \cdot 5) = 4 \cdot 49$