ABCD-isosceles trapezoid angle C = 120 degrees DC = 8 centimeters CB = 6 centimeters. Find the middle line.

Let’s draw the heights of the trapezoid CH and ВK. In a right-angled triangle СDН, the angle НСD = BCD – ВСН = 120 – 90 = 30.

Since the leg DH lies against the angle 30, its length is equal to half the length of the hypotenuse CD.

DН = СD / 2 = 8/2 = 4 cm.

Since the trapezoid is isosceles, the segment AK = DH = 4 cm.

VSNK is a rectangle, since BC is parallel to AD as the base of the trapezoid, and VK and CH are perpendicular to the bases of the trapezoid, then BC = KH = 6 cm.

Let’s calculate the length of the base AD. AD = AK + DH + KH = 4 + 4 + 6 = 14 cm.

Let’s calculate the length of the middle line of the trapezoid.

PM = (BC + AD) / 2 = (6 + 14) / 2 = 10 cm.

Answer: The middle line of the trapezoid is 10 cm.