ABCD – rectangle, angle AOB = 30, BD = 12 cm, BF perpendicular to ABC, BF = 4. Find the area ACF.

Since AВСD is a rectangle, its diagonals are equal, and the point of their intersection divides them in half.
Then ВO = DO = ВD / 2 = 12/2 = 6 cm.
From vertex B of the rectangle, draw the height BH to the AC diagonal. Then, in a right-angled triangle ВO, the leg ВН lies opposite the angle of 300, which means its length is equal to half the length of the ВO hypotenuse. ВН = ВO / 2 = 6/2 = 3 cm.
Since FB is a perpendicular in AВСD, then the BH segment is the projection of the FH segment on the plane of the rectangle, and since BH is perpendicular to the AC, then FH is perpendicular to the AC, and therefore FH is the height of the AFC triangle.
From the right-angled triangle FBH we determine the length of the hypotenuse FH.
FH ^ 2 = BH ^ 2 + FB ^ 2 = 9 + 16 = 25 cm.
FH = 5 cm.
Determine the area of ​​the triangle AFC.
S = AC * FH / 2 = 12 * 5/2 = 30 cm2.
Answer: The area of ​​the triangle is 30 cm2.