Find the area of a rectangle with a perimeter of 72 diagonals which intersect at an angle of 60.

Let us introduce the notation: a – the smaller side of the rectangle, b – the large side of the rectangle, d – its diagonal.
Consider a triangle formed by the smaller side of the rectangle and the halves of the diagonals. Because the diagonals of the rectangle are equal, then this triangle is isosceles, and the angles at the smaller side are equal to each other. The angle between the diagonals is 60, we find the angles between the smaller side and the diagonal: (180-60) / 2 = 120 * 2 = 60. If all the angles of a triangle are 60 degrees, then such a triangle is equilateral, therefore, in this rectangle, the diagonal is equal to twice the length of the smaller side: d = 2a.
The area of ​​the rectangle can be found through the diagonals: S = 0.5 * d ^ 2 * sinα = 0.5 * d ^ 2 * sin60 = 0.5 * 4a ^ 2 * √3 / 2 = √3 * a ^ 2.
Knowing that the perimeter of the rectangle is 72, you can find the half-perimeter: a + b = 72/2 = 36;
Let’s square both sides of the equality:
(a + b) ^ 2 = 36 ^ 2;
a ^ 2 + b ^ 2 + 2ab = 1296.
The sum of the squares of the adjacent sides of a rectangle is equal to the square of its diagonal, and the product of the adjacent sides is equal to the area. Then:
a ^ 2 + b ^ 2 + 2ab = d ^ 2 + 2S.
Knowing that d = 2a and S = √3 * a ^ 2, we get:
4a ^ 2 + 2 * √3 * a ^ 2 = 1296;
2a ^ 2 + √3 * a ^ 2 = 648;
a ^ 2 * (2 + √3) = 648;
a ^ 2 = 648 / (2 + √3).
Find the required area of ​​the rectangle: S = √3 * a ^ 2 = √3 * 648 / (2 + √3) ≈300.74.