Find the area of an isosceles trapezoid, the side of which is 6 cm, the lower base is 10√3

Find the area of an isosceles trapezoid, the side of which is 6 cm, the lower base is 10√3, and the angle at the base is 30 °.

Let’s build the height BH of the trapezoid ABCD.

In a right-angled triangle ABH, the leg BH lies opposite the angle 30, then its length is equal to half the length of the hypotenuse AB. BH = AB / 2 = 6/2 = 3 cm.

By the Pythagorean theorem, AH ^ 2 = AB ^ 2 – BH ^ 2 = 36 – 9 = 27.

AH = 3 * √3 cm.

Then the length DН = АD – АH = 10 * √3 – 3 * √3 = 7 * √3 cm.

Since the trapezoid is isosceles, the length of the segment DH is equal to the length of the midline of the trapezoid.

Then Savsd = DH * BH = 7 * √3 * 3 = 21 * √3 cm2.

Answer: The area of the trapezoid is 21 * √3 cm2.