The diagonals of the rhombus ABCD intersect at point O. The segment OF is the perpendicular to the side AD.

The diagonals of the rhombus ABCD intersect at point O. The segment OF is the perpendicular to the side AD. Calculate the length of the side of the rhombus if it is known that BD = 12 cm, FD = 4 cm.

The diagonals of the rhombus are halved at the intersection. Means:

OD = BD / 2 = 12/2 = 6 cm.

Consider the triangle DFO. It is rectangular, OD is the hypotenuse. Find by the Pythagorean theorem OF:

OF = √ (6² – 4²) = √ (36 – 16) = √20 = 2√5 cm.

Consider the triangle AOD. It is rectangular and the AD side is hypotenuse. Its area on one side is

S = ½ * AO * OD,

on the other side

S = ½ * AD * OF.

Let’s equalize both formulas:

½ * AO * OD = ½ * AD * OF,

AO * OD = AD * OF.

By the Pythagorean theorem

AD = √ (AO² + OD²).

Substitute this expression into the previous formula:

AO * OD = √ (AO² + OD²) * OF.

Substitute the known values ​​and solve the equation:

AO * 6 = √ (AO² + 6²) * 2√5,

AO² * 36 = (AO² + 36) * 20,

36AO² = 20AO² + 720,

36AO² – 20AO² = 720,

16AO² = 720,

AO² = 720/16,

AO² = 45,

AO = √45,

AO = 3√5 cm.

Now we can find the side of the rhombus:

AD = √ ((3√5) ² + 6²) = √ (45 + 36) = √81 = 9 cm.

Answer: the side of the rhombus is 9 cm.