In an isosceles trapezoid with an angle at the base of 60, the lateral side is equal to the smaller

In an isosceles trapezoid with an angle at the base of 60, the lateral side is equal to the smaller base and equal to 1 find the middle line.

Let us lower the heights BH and SK to the base of AD.

In a right-angled triangle ABH, we calculate the value of the angle ABH. Angle ABH = 180 – 90 – 60 = 30.

The AH leg is located opposite an angle of 30, then AH = AB / 2 = 1/2 = 0.5 cm.

Triangle CDK is equal to triangle ABH in hypotenuse and acute angle, then DK = AH = 0.5 cm.

Since BH and SK are heights, then BCKH is a rectangle, then HK = BC = 1 cm.

Base length AD = AH + NK + DK = 0.5 + 1 + 0.5 = 2 cm.

Determine the length of the midline of the trapezoid.

KP = (BC + AD) / 2 = (1 + 2) / 2 = 1.5 cm.

Answer: The length of the middle line is 1.5 cm.