In an isosceles trapezoid, a 4 cm long diagonal is formed by the base of an angle of 60 degrees. Find the midline of the trapezoid.
From the top of an obtuse angle C, we lower the height of CH to the base of AD.
Since the trapezium, by condition, is isosceles, then by the property of the height of an isosceles trapezoid drawn from the top of an obtuse angle, the height divides the larger base into two segments, the smaller of which is equal to the half difference of the bases, and more according to the sum of the bases.
AH = (AD + BC) / 2, DH = (AD – BC) / 2.
Since the middle line is also equal to the half-sum of the bases, then the segment AH = KM.
From the right-angled triangle ACH, we determine the length of the segment AH. Angle ACH = 180 – 90 – 60 = 30. Leg AH lies opposite angle 30, and therefore is equal to half the length of the hypotenuse AC.
AH = AC / 2 = 4/2 = 2 cm.
KM = AH = 2 cm.
Answer: The middle line is 2 cm.
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