The length of the side of the rhombus ABCD is 5 cm, the length of the diagonal ВD = 8 cm.

The length of the side of the rhombus AВСD is 5 cm, the length of the diagonal ВD = 8 cm. Through the point O of the intersection of the diagonals of the rhombus, a straight line OK is drawn, perpendicular to its plane. Find the distance from point K to the vertices of the rhombus if OK = 10 cm.

first we need to consider the СОВ triangle. in order to consider it, we find the OB, which is equal to half of the ВD (the diagonals in the rhombus are divided in half by the intersection point), that is, OB = 4 / now we will consider the СОВ triangle. this triangle is rectangular because the diagonals in the rhombus intersect at right angles. by the Pythagorean theorem: OC ^ 2 = BC ^ 2-OB ^ 2, that is, OC ^ 2 = 25 – 16 = 9, therefore, OB = 3 cm.
then we define the plane of the triangle KOС (through three points) and prove that KO is perpendicular to the OС (by definition, perpendicular to the straight line and plane). This means that the triangle KOС is rectangular. by the Pythagorean theorem, we will find KС: KС ^ 2 = KO ^ 2 + OC ^ 2 = 100 + 9 = 109, that is, KС = root of 109 cm = about 10.4 cm
similarly, we set the plane of the ВKС triangle, which is also rectangular. (ВO is perpendicular to KO by the same definition) and look for ВK by the Pythagorean theorem. Hence ВK = root of 41, approximately 6.4 cm.