Through the point of intersection of the diagonals of the square ABCD, a perpendicular to its plane MO

Through the point of intersection of the diagonals of the square ABCD, a perpendicular to its plane MO is drawn, equal to 2 roots of 2 cm. The side of the square is 4 cm. Find: 1) the lengths of the oblique MA, MB, MC, MD. 2) the angle between each inclined and its projection onto the plane of the square.

Since there is a square at the base, the lengths of its diagonals are equal, and at the point of intersection they are divided in half and form a right angle at the point of intersection, then AO = BO = CO = DO, then the lengths of oblique MA = MB = MC = MD.

It is enough to find the length of one oblique.

Consider a right-angled triangle of COB, in which the hypotenuse BC = 2 cm, and the legs BO and CO are equal. Then, by the Pythagorean theorem, CB ^ 2 = 2 * OB ^ 2.

ОВ ^ 2 = CB ^ 2/2 = 16/2 = 8.

ОВ = 2 * √2 cm.

Consider a right-angled triangle MOS, and by the Pythagorean theorem, we determine the length of the hypotenuse MC.

MC ^ 2 = OС ^ 2 + OM ^ 2 = (2 * √2) ^ 2 + (2 * √2) ^ 2 = 8 * 8 = 16.

MС = √16 = 4 cm.

MС = MA = MB = MD = 4 cm.

In the OMC triangle, the leg OС = OM = 2 * √2, then the triangle is isosceles and right-angled, then the angle OCM = 45.

The angles between other oblique and oblique projections are also 45.

Answer: MA = MВ = MС = MD = 4 cm.The angles between the oblique and their projections are 45.