The base side of a regular quadrangular prism is 10. The diagonal of the parallelepiped is inclined

The base side of a regular quadrangular prism is 10. The diagonal of the parallelepiped is inclined to the base plane at an angle of 60 degrees. Find the area of the lateral surface of the parallelepiped.

Let ABCDA₁B₁C₁D₁ be a regular quadrangular prism.
Since it is correct, it means that at the base there is a square (ABCD) with a side = 10
∠С₁АС = 60 °
Find the diagonal in the square ABCD according to the Pythagorean theorem:
AC = √ (10 ^ 2 + 10 ^ 2) = √ (100 + 100) = √200 = 10√2
Now consider ΔС₁АС:
∠АС₁С = 180 ° -90 ° -60 ° = 30 °
By the property of a right-angled triangle: a leg lying opposite an angle of 30 ° = half of the hypotenuse.
Therefore, АС₁ = 2АС = 2 * 10√2 = 20√2
It is rectangular, so by the Pythagorean theorem one can find CC₁
CC₁ = √ (20√2) ^ 2 – (10√2) ^ 2) = √ (400 * 2-100 * 2) = √ (800-200) = √600 = 10√6
Now you can find the area of ​​the side surface of the parallelepiped.
Since the parallelepiped is regular, all its side faces are equal.
And the area of ​​the side surface of a parallelepiped is the sum of the areas of the side faces.
Therefore, S = (10 * 10√6) * 4 = 100√6 * 4 = 400√6
Answer: 400√6