ABCD-rhombus, side 4, angle D = 150 degrees, BM-perpendicular to the plane

| August 10, 2021 | education

ABCD-rhombus, side 4, angle D = 150 degrees, BM-perpendicular to the plane of the rhombus and it is equal to 2√3. Find the distance from point M to AD.

From the top B we draw the height BH to the AD side of the rhombus.

The BH segment is the projection of the MH segment on the ABC plane. Then the segment MH is perpendicular to the side AD, which is the shortest distance from point M to the side AD.

In a rhombus, the opposite angles are equal, then the angle ABC = ADC = 150.

Since the angle HBD = 90, then the angle ABH = ABC – 90 = 60, then the angle BAH = 180 – 90 – 60 = 30.

Leg BH, right-angled triangle ABH lies opposite angle 30, then BH = AB / 2 = 4/2 = 2 cm.

In the right-angled triangle MBH, we determine the length of the hypotenuse MH, applying the Pythagorean theorem.

MH ^ 2 = MB ^ 2 + BH ^ 2 = (2 * √3) ^ 2 + 2 ^ 2 = 12 + 4 = 16.

MH = 4 cm.

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



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