One of the corners of the rhombus is 120 degrees and the smaller diagonal is 4.5 Find the perimeter of the rhombus.

Let us denote the vertices of the rhombus by the letters A, B, C, D. By the letter O we denote the point of intersection of the diagonals.
DAB angle = 120 °. Hence, the angle OAB = 60 °, since the AC diagonal divides the angle in half.
Since our rhombus is divided into right-angled triangles, consider the triangle OAB.
We know that the angle OAB = 60 °. Hence the angle ABO = 30 °.
Since the diagonal of the rhombus is divided in half at the point of intersection, we have AO = 0.5 AC. We get AO = 0.5 * 4.5 = 2.25 cm.
Opposite the 30 ° angle lies the leg. which is half the hypotenuse.
If AO = 2.25 cm, then AB, being the hypotenuse of a right-angled triangle, will be equal to 2 * AO
AB = 2 * 2.25 = 4.5 cm.
We know that all sides of a rhombus are equal.
The perimeter of the rhombus will be P = 4 * AB, Z = 4 * 4.5 cm = 18 cm.
Answer: The perimeter of the rhombus is 18 cm.