Find the midline of a rectangular trapezoid ABCD if its angle D is 30

| May 23, 2021 | education

Find the midline of a rectangular trapezoid ABCD if its angle D is 30, and the diagonal AC is perpendicular to the side of CD and the base of AD is 16.

The ACD triangle is rectangular, since, by condition, AC is perpendicular to CD. The leg AC lies opposite the angle 30, then AC = AD / 2 = 16/2 = 8 cm, then, according to the Pythagorean theorem, we determine the length of the leg CD.

CD ^ 2 = AD ^ 2 – AC ^ 2 = 256 – 64 = 192.

СD = 8 * √3 cm.

Let’s build the height of the CH trapezoid ABD.

The CDH triangle is rectangular, then the CH leg lies opposite the angle 30, then CH = CD / 2 = 8 * √3 / 2 = 4 * √3 cm.

DH ^ 2 = CD ^ 2 – H ^ 2 = 192 – 48 = 144.

DН = 12 cm.

Then AH = BC = 12 cm.

Let’s define the middle line of the trapezoid.

KM = (AD + BC) / 2 = (16 + 12) / 2 = 14 cm.

Answer: The middle line of the trapezoid is 14 cm.



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