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.
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