To the plane of the rhombus ABCD in which LA = 45 °, AB = 8 cm, a perpendicular

To the plane of the rhombus ABCD in which LA = 45 °, AB = 8 cm, a perpendicular MС with a length of 7 cm is drawn. Find the distance from point M to line AB.

Let us determine the height DH to the side AB of the rhombus ABCD.

Since the angle BAD, by condition, is equal to 45, then the triangle ADH is rectangular and isosceles. AH = DH = AB * Sin45 = 8 * √2 / 2 = 4 * √2 cm.

From the vertex C we construct the height of the SC for the continuation AB.

SK = DH = 4 * √2 cm.

The MSC triangle is rectangular, in which, according to the Pythagorean theorem, MK ^ 2 = CK ^ 2 + MC ^ 2 = 32 + 49 = 81.

MK = 9 cm.

Answer: From point M to straight AB 9 cm.