The length of the side of the rhombus ABCD is 5 cm, the length of the diagonal BD is 6 cm. Through the point O of the intersection of the diagonal of the rhombus, a straight line OK is drawn, perpendicular to its plane. Find the distance from point K to the top of the rhombus if OK = 8 cm.
The diagonals of the rhombus, at the point of their intersection, are divided in half and intersect at right angles, then BO = BD / 2 = 6/2 = 3 cm, and triangle AOB is rectangular.
In a right-angled triangle AOB, according to the Pythagorean theorem, AO ^ 2 = AB ^ 2 – OB ^ 2 = 25 – 9 = 16. OA = CO = 4 cm.
Since OK is perpendicular to the plane of the rhombus, the triangles KOA and KOC are rectangular.
Then KB ^ 2 = KO ^ 2 + OB ^ 2 = 64 + 9 = 73, KC ^ 2 = KO ^ 2 + OC ^ 2 = 64 + 16 = 80.
КВ = √73 cm, КС = √80 = 4 * √5 cm.
Answer: From point K to the tops of the rhombus √73 cm and 4 * √5 cm.
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