One of the corners of the rhombus is 60 degrees and the diagonal drawn from

One of the corners of the rhombus is 60 degrees and the diagonal drawn from the top of this corner is 4 cm, find the perimeter of the rhombus.

Since ABCD is a rhombus, the lengths of its sides are equal, AB = BC = CD = AD, then triangle ABD is isosceles, and since the angle BAD, by condition, is 60, then triangle ABD is equilateral.

The diagonals of the rhombus at point O are halved and intersect at right angles, then AO = AC / 2 = 4/2 = 2 cm, and AO is the height, bisector and median of triangle ABD.

The height of an equilateral triangle is: AO = AB * √3 / 2.

AB = 2 * AO / √3 = 4 / √3 = 4 * √3 / 3 cm.

Then Ravsd = 4 * AB = 4 * 4 * √3 / 3 = 16 * √3 / 3 cm.

Answer: The perimeter of the rhombus is 16 * √3 / 3 cm.