In a rectangular trapezoid ABCD, the acute angle BAD is 60˚. Find the length of the midline of the trapezoid if AB = AD = 8cm.

| March 5, 2021 | education

Given: ABCD – rectangular trapezoid; angle BAD = 60 degrees; AB = AD = 8 cm.
Find: the length of the midline of the trapezoid ABCD.
Decision:
ABCD is an isosceles trapezoid by condition. Let’s draw the height of the HV. The resulting triangle BHA is rectangular (since the angle BHA = 90 °). If the angle BHA = 90 °, the angle BAH = 60 °, then the angle HBA = 90 ° – 60 ° = 30 ° (in a right-angled triangle, the angles at the base are 90 ° in total).
Opposite an angle of 30 ° lies a leg equal to 1/2 of the hypotenuse. In this case, the leg opposite to the 30 ° angle is AH, and the hypotenuse is AB.
By the condition AB = AD = 8. So AH = 1/2 x AB = 1/2 x 8 = 8/2 = 4 cm. AH = 4 cm.
DH = AD – AH = 8 – 4 = 4 cm. Hence, the base CB = 4 cm (since, having drawn the height, we have a rectangle DCBH, in which CB = DH). Base AD = 8 cm.
The middle line of the trapezoid is equal to the half-sum of its bases: 1/2 x (CB + AD) = 1/2 x (4 + 8) = 1/2 x 12 = 12/2 = 6 cm.
Answer: The middle line of the trapezoid is 6 cm.



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