The base of an oblique parallelepiped is a rhombus with a side length of 60. The plane of the diagonal section passing through the larger diagonal of the base is perpendicular to the plane of the base. The area of this section is 7200. Find the smaller diagonal of the base, if the lateral edge is 80 and forms an angle of $60^{2}$ with the base plane.


#
Let the plane of the diagonal section $A A1 C1 C$, passing through the larger diagonal $AC$ of the base $ABCD$ of the parallelepiped $ABCD A1 B1 C1 D1$, be perpendicular to the base $ABCD$. Then the perpendicular $C1K$, dropped from the vertex $C1$ to the plane of the base $ABCD$, is the height of the parallelepiped $ABCD A1 B1 C1 D1$.

Thus, $C1CK$ is the angle of the lateral edge $CC1$ with the plane of the base $ABCD$. According to the problem, $\angle C1CK=60^{\circ}$. In addition,

$$
S_{A}1 C1 C=CC1 \cdot AC \sin \angle C1CK=80 AC \cdot \sin 60^{\circ}=40 AC \cdot \sqrt{3}=7200
$$

from which we find that $AC=60 \sqrt{3}$. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides, so

$$
BD^2 + AC^2 = 4 \cdot AB^2, \quad
$$

from which we find that

$$
BD^2 = 4 \cdot AB^2 - AC^2 = 4 \cdot 60^2 - 3 \cdot 60^2 = 60^2
$$

Therefore, $BD=60$.

## Answer

60.00