# ABCD is a rhombus with a side of 4 cm. Angle ADC 150. BM is a perpendicular to the plane of the rhombus

**ABCD is a rhombus with a side of 4 cm. Angle ADC 150. BM is a perpendicular to the plane of the rhombus and is equal to 2√3. find the distance from point M to AD.**

Let us draw a perpendicular from the vertex B of the rhombus to the side AD. Since ВM is perpendicular to the plane of the rhombus, and ВН is perpendicular to AD, along the segment ВН is perpendicular to AD and is our desired distance.

The sum of the adjacent angles of the rhombus is 180, then the angle BAD = 180 – 150 = 30. In the right-angled triangle ABН, the leg BН lies opposite the angle 30, then its length is equal to half the length of the hypotenuse AB. ВН = AB / 2 = 4/2 = 2 cm.

In a right-angled triangle ВMН, according to the Pythagorean theorem, the hypotenuse of MН will be equal to: MH ^ 2 = BM ^ 2 + BH ^ 2 = (2 * √3) ^ 2 + 2 ^ 2 = 12 + 4 = 16.

MH = √16 = 4 cm.

Answer: The distance from point M to AD is 4 cm.