ABCD-square with a side of 4 cm. The AMB triangle has a common side AB with a square, AM = BM

ABCD-square with a side of 4 cm. The AMB triangle has a common side AB with a square, AM = BM = 2√6 cm. The planes of the triangle and the square are mutually perpendicular. 1) Prove that BC is perpendicular to AM. 2) Find the angle between the MC and the plane of the square. 3) Find the distance from A to the DMC plane.

By condition, the plane of the triangle ABM is perpendicular to the plane of the square ABCD.

ABCD square, then AD is parallel to BC. Segments AM and AB leaving the same point of perpendicular planes, then AM is perpendicular to AD, and since BC is parallel to AD, then AM is perpendicular to BC, which was required to be proved.

Let us construct the height of the MH of the isosceles triangle ABM. The segment CH is the projection of the inclined CM onto the plane of the square, then the linear angle of the MCH is the angle between the MS and the plane of the square.

Divide point H AB in half, then AH = AB / 2 = 4/2 = 2 cm.

By the Pythagorean theorem, MH ^ 2 = AM ^ 2 – AH ^ 2 = 24 – 4 = 20. MH = √20 cm.

In a right-angled triangle BCH, according to the Pythagorean theorem, CH ^ 2 = BC ^ 2 + BH ^ 2 = 16 + 4 = 20.

CH = √20 cm.Then the triangle MCH is rectangular and isosceles, and the angle MCH = 45.

Through point M we draw a straight line PM parallel to CD. Let us drop the perpendicular AP from point P.

AR = MH = √20 cm.

Triangle APD is rectangular, in which the distance AK is the distance from point A to the plane DMC.

Let us determine the length of the hypotenuse DP.

DP ^ 2 = AP ^ 2 + AD ^ 2 = 20 + 16 = 36.

DP = 6 cm.

Then AK = AR * AD / DP = √20 * 4/6 = 2 * √5 * 4/6 = 4 * √5 / 3 cm.

Answer: The angle between the MC and the plane of the square is 45, from point A to the DMC plane 4 * √5 / 3 cm.