The diagonals of the rhombus ABCD intersect at point O. On the side AB, point K is taken so that OK is perpendicular

The diagonals of the rhombus ABCD intersect at point O. On the side AB, point K is taken so that OK is perpendicular to AB, Ak = 8 cm, BK = 8 cm. Find the diagonals of the rhombus.

The diagonals of the rhombus, at the point of their intersection, are halved and intersect at right angles. Then triangle AOB is rectangular, AO = CO, BO = DO.

In a right-angled triangle AOB, the height OK is divided in half by the hypotenuse AB, and then OK and the median of the triangle, and then the triangle AOB is isosceles.

Since OA = OB, then the quadrangle ABCD is a square, then AC = BD.

By the Pythagorean theorem, AO ^ 2 + BO ^ 2 = AB ^ 2.

2 * AO ^ 2 = 256.

AO = 16 / √2 = 8 * √2 cm.

Then AC = BD = 2 * AO = 16 * √2 cm.

Answer: The diagonals are 16 * √2 cm.