Find the perimeter of the rhombus ABCD, in which the angle B = 60 degrees, AC = 10.5 cm.

The perimeter of a rhombus is the sum of the lengths of all its sides. Since in a rhombus all sides are equal, the perimeter of a rhombus is found by the following formula: P = 4 * AB (AB is the side of the rhombus).

The diagonal of a rhombus is the bisector of its angles (according to the property of the diagonals of a rhombus). Thus, the diagonal AC is the bisector ∠ A and ∠ C. Given that the opposite angles of the rhombus are equal, ∠ BAC = ∠ BCA.

Consider a triangle BАС:
∠ АBС = 60 °;
∠ BАС = ∠ ВСА.
Since the sum of the angles of a triangle is 180 °, ∠ ABC + ∠ BAC + ∠ BCA = 180 °.
60 ° + 2 * ∠ BAC = 180 °;
2 * ∠ BAC = 180 ° – 60 °;
2 * ∠ BAC = 120 °;
∠ BAC = 120 °: 2;
∠ BAC = ∠ BAC = 60 °.

Because in the triangle BAC ∠ ABC = ∠ BAC = ∠ BCA = 60 °, triangle BAC is equilateral, therefore AB = BC = AC = 10.5.

Find the perimeter of the rhombus ABCD.
P = 4 * 10.5 = 42.

Answer: 42.