ABCD is a rhombus with a side equal to a. Angle A = 60 degrees, MA⊥ ABC, MA = a / 2. Find the distance from point M along the straight line DC.
Let’s draw the height АН to the side of the rhombus CD.
Since AM is perpendicular to the rhombus plane, AM is perpendicular to AH. АН is the projection of МН onto the rhombus plane, then МН is a perpendicular to СН.
The angle BAН = 90, then the angle DAН = 90 – 60 = 30. Then the leg DH = AD / 2 = a / 2, since it lies opposite the angle 30.
In the right-angled triangle MAН, we define the length of the hypotenuse MH.
MH ^ 2 = AM ^ 2 + AH ^ 2 = (a / 2) ^ 2 + (a * √3 / 2) ^ 2 = a ^ 2/4 + 3 * a ^ 2/4 = a ^ 2 cm2.
Answer: The distance from point M to СD is a2 cm2.
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