10,11

In the parallelepiped $A B C D A 1 B 1 C 1 D 1$, the face $A B C D$ is a square with side 5, the edge $A A 1$ is also 5, and this edge forms angles of $60^{\circ}$ with the edges $A B$ and $A D$. Find the diagonal $B D 1$.
## First Method

Triangle $A A 1 B$ is equilateral, since $A A 1 = A B$ and $\angle B A A 1 = 60^{\circ}$. Therefore, $A 1 B = A A 1 = 5$. Similarly, $A 1 D = 5$. The lateral edges $A 1 A, A 1 B$, and $A 1 D$ of the triangular pyramid $A 1 A B D$ with vertex $A 1$ are equal, so the height $A 1 O$ of this pyramid passes through the center of the circle circumscribed around the base $A B D$. Since triangle $A B D$ is a right triangle, point $O$ is the midpoint of its hypotenuse $B D$, i.e., the center of the square $A B C D$. From the right triangle $O B A 1$, we find that

$$
A 1 O = \sqrt{A_{1} B^{2} - B O^{2}} = \sqrt{5^{2} - \left(\frac{5}{\sqrt{2}}\right)^{2}} = \frac{5}{\sqrt{2}}
$$

Since $D 1 C = A 1 B = A 1 A = D 1 D$, point $D 1$ is equidistant from vertices $C$ and $D$, so its orthogonal projection $K$ on the plane of the base $A B C D$ is also equidistant from $C$ and $D$, and thus lies on the perpendicular bisector of segment $C D$. Since $D 1 K \parallel A 1 O$ and $D 1 K = A 1 O$, quadrilateral $A 1 D 1 K O$ is a rectangle, so $O K = A 1 D 1 = 5$. Extend segment $K O$ to intersect segment $A B$ at point $M$. Then $M$ is the midpoint of $A B$ and $M K = M O + O K = \frac{15}{2}$. From the right triangles $M K B$ and $K B D 1$, we find that

$$
\begin{gathered}
B K = \sqrt{B M^{2} + M K^{2}} = \sqrt{\frac{25}{4} + \frac{225}{4}} = \sqrt{\frac{250}{4}} \\
B D 1 = \sqrt{K D_{1}^{2} + B K^{2}} = \sqrt{\frac{50}{4} + \frac{250}{4}} = \sqrt{\frac{300}{4}} = \sqrt{75} = 5 \sqrt{3} .
\end{gathered}
$$

Second Method.

$$
\begin{gathered}
\overrightarrow{B D_{\mathbf{1}}} = \overrightarrow{B A} + \overrightarrow{A A_{1}} + \overrightarrow{A_{\mathbf{1}} D_{\mathbf{1}}} \Rightarrow B D 12 = \left(\overrightarrow{B A} + \overrightarrow{A A_{\mathbf{1}}} + \overrightarrow{A_{\mathbf{1}} D_{\mathbf{1}}}\right)^2 = \\
= \overrightarrow{B A}^2 + \overrightarrow{A A_{\mathbf{1}}} + \overrightarrow{A_{\mathbf{1}} D_{\mathbf{1}}} + 2 \overrightarrow{B A} \cdot \overrightarrow{A A_{\mathbf{1}}} + 2 \overrightarrow{B A} \cdot \overrightarrow{A_{\mathbf{1}} D_{\mathbf{1}}} + 2 \overrightarrow{A A_{\mathbf{1}}} \cdot \overrightarrow{A_{\mathbf{1}} D_{\mathbf{1}}} = \\
= 25 + 25 + 25 + 2 \cdot 25 \cos 120^{\circ} + 2 \cdot 25 \cos 90^{\circ} + 2 \cdot 25 \cos 60^{\circ} = 75 \Rightarrow \\
\Rightarrow B D 1 = \sqrt{75} = 5 \sqrt{3} .
\end{gathered}
$$

## Answer

$5 \sqrt{3}$.