# Through the point O of the intersection of the diagonals of a square with a side of 4 cm, a straight line OM

**Through the point O of the intersection of the diagonals of a square with a side of 4 cm, a straight line OM is drawn, perpendicular to the plane of the square. Find the distance from point M to the top of the square if OM = 2√2cm.**

From the properties of the square and its diagonals, we know that the sides of the square are equal, the angles are straight, the diagonals are equal, mutually perpendicular and divide each other in half.

Since the diagonal and the two sides of the square form a right-angled triangle, based on the Pythagorean theorem, we find the length of the diagonal of the square (we denote it by “c”):

c ^ 2 = a ^ 2 + b ^ 2 = 4 ^ 2 + 4 ^ 2 = 16 + 16 = 32;

c = √‾32 = √‾ (2 * 16) = 4 * √‾2.

Half of the diagonal of the square (the segment from any vertex of the square to point O) is equal to:

s / 2 = 4 * √‾2 / 2 = 2 * √‾2.

The triangle formed by half the diagonal of the square, the perpendicular OM and the segment from M to the apex of the square is also rectangular and its legs are known. The segment from point M to the apex of the square is the hypotenuse of this triangle, we find its length:

√‾ ((s / 2) ^ 2 + [OM] ^ 2) = √‾ ((2 * √‾2) ^ 2 + (2 * √‾2) ^ 2) = √‾ (4 * 2 + 4 * 2) = √‾ (8 + 8) = √‾ (16) = √‾ (42) = 4.