In the rectangular parallelepiped ABCDA1B1C1D1, a section plane is drawn containing the diagonal AC1, so that the section

In the rectangular parallelepiped ABCDA1B1C1D1, a section plane is drawn containing the diagonal AC1, so that the section is a rhombus. Find the area if AB = 3, BC = 2 and AA1 = 5

AC1 is the larger diagonal D of the diamond.
AA1 = BB1 = DD1 = CC1 = 5, AD = A1D1 = CB = C1B1 = 2, DC = D1C1 = AB = A1B1 = 3.
Rhombus area:
S = ½ * D * d, d is the smaller diagonal.
d = (4 * a ^ 2 – D ^ 2) ^ (1/2), where a is the side of the rhombus.
From right-angled triangle ACD:
AC ^ 2 = AD ^ 2 + DC ^ 2 = 9 + 4 = 13.
The AC1C triangle is rectangular, therefore
AC1 = D = (AC ^ 2 + C1C ^ 2) ^ (1/2) = (13 + 25) ^ (1/2) = √38.
Consider right triangle AMD, M belongs to DD1.
Let MD = x.
Then,
a ^ 2 = AD ^ 2 + MD ^ 2 = 4 + x ^ 2.
On the other hand, from right-angled triangle D1C1M:
a ^ 2 = C1D1 ^ 2 + D1M ^ 2 = 9 + (5 – x) ^ 2.
Then,
4 + x ^ 2 = 9 + (5 – x) ^ 2.
4 + x ^ 2 = 9 + 25 – 10 * x + x ^ 2.
X = 3.
Hence,
a ^ 2 = 4 + 9 = 13.
d = (4 * a ^ 2 – D ^ 2) ^ (1/2) = (4 * 13 – 38) ^ (1/2) = √14.
S = ½ * D * d = ½ * √38 * √14 = ½ * 2 * √133 = √133.