In rhombus ABCD, angle B-angle A = 60 degrees. Segments BF and BT are bisectors of triangles ABD and BDC

In rhombus ABCD, angle B-angle A = 60 degrees. Segments BF and BT are bisectors of triangles ABD and BDC, respectively. Calculate the area of a rhombus if the distance from point F to line BF is 4 cm.

Since all sides of a rhombus are equal, triangle ABD is isosceles, and since angle A = 60, this triangle is equilateral. Then the bisectors ВF and ВТ are the same as the heights of the triangles.

In a rhombus, the sum of adjacent angles is 180, then the ACD angle = 180 – 60 = 120.

In the quadrangle FВТD the angle F and T are equal to 90, then the angle FВТ = 360 – 90 – 90 – 120 = 60.

Since the rhombus has the same heights, the BTF triangle is equilateral.

The height of an equilateral triangle will be: FH = a * √3 / 2, where a is the side of the triangle.

BT = FH * 2 / √3 = 2 * 4 / √3 = 8 / √3 cm.

In an equilateral triangle ABD, the height is BF = 8 / √3 cm, then AB = BF * 2 / √3 = (8 / √3) * 2 / √3 = 16/3.

Determine the area of ​​the rhombus. Savsd = AB * BF = (16/3) * (8 / √3) = 128/3 * √3 = 128 * √3 / 9 cm3.

Answer: The area of ​​the rhombus is 128 * √3 / 9 cm3.