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.
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