In rhombus ABCD, angle A is equal to 60 degrees AB 6 cm from vertex B to sides ABCD, perpendiculars BM

In rhombus ABCD, angle A is equal to 60 degrees AB 6 cm from vertex B to sides ABCD, perpendiculars BM and BK are drawn, respectively, which is equal to the sum of the lengths of the segments if 4 cm.

Since ABCD is a rhombus, the lengths of all its sides are equal, as well as the opposite angles at the vertices.

AB = BC = CD = AD = 6 cm, angle BAC = BCD = 60.

Then the triangle ABD is isosceles, since AB = AD, and since the angle at the apex is A = 60, the triangle ABD is equilateral.

Similarly, the BCD triangle is equilateral.

Then the heights BM and BK are also the medians of the triangles ABD and BCD, then AM = DM = AD / 2 = 6/2 = 3 cm, СK = DK = CD / 2 = 6/2 = 3 cm.

DМ + DK = 3 + 3 = 6 cm.

Answer: The sum of the segments DM and DK is 6 cm.