From the vertex of the square ABCD, the perpendicular AE to the plane of the square is restored

From the vertex of the square ABCD, the perpendicular AE to the plane of the square is restored, which is equal to the distance from E to BD, if AE is 2dm, AB-8.

Let’s construct the diagonals AC and BD of the square ABCD.

The diagonals AC and BD intersect at right angles and are halved at the point of intersection. The segment OA is the projection of the inclined OE on the plane of the square, and since AO is perpendicular to BD, then OE is perpendicular to BD.

Triangle AOB is rectangular and isosceles, then AB ^ 2 = 2 * AO ^ 2.

AO ^ 2 = AB ^ 2/2 = 64/2 = 32.

In a right-angled triangle AOE, according to the Pythagorean theorem, OE ^ 2 = AO ^ 2 + AE ^ 2 = 32 + 4 = 36. OE = 6 cm.

Answer: From point E to segment BD 6 cm.