In an isosceles trapezoid, the acute angle is 60 degrees, and the lateral side is 10 cm

In an isosceles trapezoid, the acute angle is 60 degrees, and the lateral side is 10 cm, the smaller base is 14 cm. Find the midline of the trapezoid.

Let’s call the trapezoid ABCD: AB, CD – sides, BC, AD – bases. AB = BC = 10 cm. Draw the height BK, then consider the triangle AKB (angle K = 90):
cos (KAB) = AK / AB;
AK = AB * cos (KAB);
AK = 10 * (1/2) = 5 cm.
Since the length BC = 14 cm, then we draw another height CL, length LD = AK = 5 cm, length KL = 14 cm.
Then the length of AD:
AD = KL + AK + LD = 14 + 5 + 5 = 24 cm.
Then the middle line of the trapezoid has a length:
(BC + AD) / 2 = (14 + 24) / 2 = 19 cm.
Answer: the length of the middle line is 19 cm.