10,11

In the pyramid $ABCD$, the edges are given: $AB=7, BC=8, CD=4$. Find the edge $DA$, if it is known that the lines $AC$ and $BD$ are perpendicular.
Let's draw the height $B K$ of triangle $A B C$. The line $A C$ is perpendicular to two intersecting lines $B K$ and $B D$ in the plane $B D K$. Therefore, the line $A C$ is perpendicular to each line in this plane, particularly to the line $D K$. Hence, $D K$ is the height of triangle $A D C$. We can apply the Pythagorean theorem to the right triangles $A B K$, $C B K$, $A D K$, and $C D K$:

$$
A B^2 - A K^2 = B C^2 - C K^2, \quad A D^2 - A K^2 = D C^2 - C K^2.
$$

Subtracting these equations term by term, we get the equation

$$
A B^2 - A D^2 = B C^2 - D C^2, \text{ or } 49 - A D^2 = 64 - 16
$$

from which we find that $A D^2 = 1$. Therefore, $A D = 1$.

## Answer

1.