In the MNKT rhombus, the angle N is 60 degrees, MK = 10.5 cm. Find the perimeter of the rhombus.

The NOK triangle is rectangular, the KON angle is 90 degrees (the diagonals of the rhombus are perpendicular).

O is the point of intersection of the diagonals, which means that KO = MO (the point of intersection of the diagonals divides the diagonal in half).

MK = 10.5 (by condition), therefore KO = 10.5 / 2 = 5.25 cm.

The angle KNO is equal to half of the angle KNM (the bisector NO bisects the angle of the rhombus N in half).

The angle KNM = 60 degrees (by condition), so the angle KNO = 60/2 = 30 degrees.

Sine of angle KNO = KO / NK. Let us denote NK for X.

The equation turns out:

Sine 30 degrees = 5.25 / x

The sine of 30 degrees is 1/2, substitute it into the equation and solve.

1/2 = 5.25 / x

x = 5.25 * 2

x = 10.5, that is, the side of the rhombus is 10.5 cm.

Find the perimeter of the rhombus
Since all sides of the rhombus are equal, then:

P = 10.5 * 4 = 42 cm.

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