# In trapezoid ABCD, the base of AD is twice as large as the base of BC

In trapezoid ABCD, the base of AD is twice as large as the base of BC and twice as large as the lateral side of CD. The ADC angle is 60 *, the AB side is 6. Find the area of the trapezoid.

From the vertices of the trapezoid B and C, we omit the heights BM and CH.

We denote the length of the base of the BC through X cm, then, by condition, CD = X cm, AD = 2 * X cm, and MH = X, since MBCH is a rectangle.

In a right-angled triangle СНD, the angle С = 180 – 90 – 60 = 30, then the leg НD lies opposite the angle 30, which means it is equal to half of the СD, НD = СD / 2 = X / 2.

Let us determine the length of the segment AM.

AM = AD – MH – HD = 2 * X – X – X / 2 = X / 2.

AM = HD = X / 2. Since the heights are cut off at the base of the segments of equal length, the trapezoid ABCD is isosceles, and AB = CD = 6 cm.

X = 6 cm, then BC = 6 cm, and AD = 2 * 6 = 12 cm.

From the triangle СНD we determine the length of the height СН according to the Pythagorean theorem, taking into account the fact that НD = X / 2 = 3.

CH ^ 2 = CD ^ 2 – HD ^ 2 = 36 – 9 = 27.

CH = √27 = 3 * √3 cm.

Then the area of ​​the trapezoid is:

S = (BC + AD) * CH / 2 = (6 + 12) * 3 * √3 / 2 = 27 * √3 cm.

Answer: The area of ​​the trapezoid is 27 * √3 cm.

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