# 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.