The diagonals of the rhombus ABCD intersect at point O, BD = 16 cm. On the side AB, point K is taken so that OK

| July 7, 2021 | education

The diagonals of the rhombus ABCD intersect at point O, BD = 16 cm. On the side AB, point K is taken so that OK is perpendicular to AB and OK = 4√3. Find the side of the rhombus and the second diagonal.

The diagonals of the rhombus are halved at the point of intersection. OB = BD / 2 = 16/2 = 8 cm.

Since OK is perpendicular to AB, the triangle OBK is rectangular.

Then, by the Pythagorean theorem, BK ^ 2 = OB ^ 2 – OK ^ 2 = 64 – 48 = 16.

BK = 4 cm.

OK is the height drawn to the hypotenuse, then OK ^ 2 = BK * AK.

AK = OK ^ 2 / BK = 48/4 = 12 cm.

Then AB = AK + BK = 12 + 4 = 16 cm.

In a right-angled triangle AOB, according to the Pythagorean theorem, AO ^ 2 = AB ^ 2 – OB ^ 2 = 256 – 64 = 192.

AO = 8 * √3 cm, then AC = 2 * AO = 16 * √3 cm.

Answer: The length of the side is 16 cm, the diagonal is 16 * √3 cm.



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