# In an isosceles trapezoid, the base is 7 cm, the side is 6 cm and 1 of the angles is 150 degrees.

**In an isosceles trapezoid, the base is 7 cm, the side is 6 cm and 1 of the angles is 150 degrees. Find the area of the trapezoid.**

In the right-angled triangle ABН, we determine the values of the acute angles.

Angle ABH = ABC – СBH = 150 – 90 = 60, then the angle BAН = 90 – 60 = 30.

The BH leg is located opposite an angle of 30, then H = AB / 2.

Sin60 = AH / AB.

AH = AB * Sin60 = 6 * √3 / 2 = 3 * √3 cm.

Let’s draw the second height of the CК, then the right-angled triangle CDK is equal to the triangle ABH along the hypotenuse and acute angle, then DK = AH = 3 * √3 cm.

НC = BC = 7 cm, since the quadrangle BCKН is a rectangle.

AD = (AН + НK + DK) = (3 * √3 + 7 + 3 * √3) = 7 + 6 * √3 cm.

Determine the area of the trapezoid.

Savsd = (BC + AD) * CH / 2 = (7 + 7 + 6 * √3) * 3/2 = 21 + 9 * √3 = 9 * (3 + √3) cm2.

Answer: The area of the trapezoid is 9 * (3 + √3) cm2.