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

| April 28, 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 at the point of their intersection are halved and intersect at right angles. Then BO = BD / 2 = 16/2 = 8 cm.

In a right-angled triangle BOK, according to the Pythagorean theorem BK ^ 2 = BO ^ 2 – OK ^ 2 = 64 – 48 = 14. BK = 4 cm.

OK – the height of a right-angled triangle, drawn to the hypotenuse, then OK ^ 2 = BK * AK.

AK = OK ^ 2 / BK = 48/4 = 12 cm. AB = AK + BK = 12 + 4 = 16 cm.

In a right-angled triangle AOB, AO ^ 2 = AB ^ 2 – OB ^ 2 = 256 – 64 = 144.

AO = 12 cm, then AC = 2 * AO = 24 cm.

Answer: The side of the rhombus is 16 cm, the second diagonal is 24 cm.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.