Find the perimeter of a rhombus if the length of its smaller diagonal is 7 cm and one of its corners is 60 degrees.

Let ABCD be a rhombus. AC and BD = 7 cm are diagonals intersecting at point O, ∠A = 60 °.
1. The diagonals of the rhombus intersect at right angles and are the bisectors of the corners of the rhombus, thus the diagonals divide the rhombus into 4 equal right-angled triangles.
Since the diagonals are halved by the intersection point, then:
BO = DO = BD / 2 = 7/2 = 3.5 (cm).
Consider a triangle AOB: AO and BO = 3.5 cm – legs, AB – hypotenuse, ∠BOA = 90 °, ∠OAB = ∠A / 2 = 60 ° / 2 = 30 °.
The VO leg lies opposite to ∠OAB = 30 °, and from the properties of a right-angled triangle it is known that the leg opposite to an angle of 30 ° is equal to half the hypotenuse, then:
BO = AB / 2;
AB / 2 = 3.5;
AB = 2 * 3.5;
AB = 7 cm.
2. All sides of the rhombus are equal:
AB = BC = CD = AD.
The perimeter of the rhombus is:
P = 4 * AB = 4 * 7 = 28 (cm).
Answer: P = 28 cm.