One of the corners of the rhombus is 120 degrees. Find the P rhombus if the diagonal of this corner outgoing

One of the corners of the rhombus is 120 degrees. Find the P rhombus if the diagonal of this corner outgoing from the vertex is 10 cm.

Let the angle ABC = 120 ° in the rhombus ABCD and the diagonal BD = 10 cm.

It is necessary to calculate the perimeter of the rhombus P.

Based on the properties of the rhombus, the ADC angle is equal to the ABC angle and is 120 °.

In addition, the diagonal BD divides the angle ABC in half.

It turns out that the angles ABD and CBD are equal to 60 °, and the angles ADB and CDB are also equal to 60 °.

It follows that the triangles ABD and CBD are regular, whence it follows:

AB = BD = AD, BC = CD = BD.

Thus, all sides of the rhombus are equal to its diagonal BD.

It turns out, P = 4 * BD.

P = 4 * 10 cm.

P = 40 cm.

Answer: the perimeter of the rhombus is 40 cm.