Find the perimeter and area of an isosceles trapezoid with an angle of 120 degrees if its bases are 3 cm and 5 cm.

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

Since the trapezoid is isosceles, the BH height divides the base into two segments, the length of the smaller of which is equal to the half-difference of the base lengths.

AH = (BP – BC) / 2 = 2/2 = 1 cm.

In a right-angled triangle ABH, the angle ABH = ABC – СBH = 120 – 90 = 30.

The AН leg is erect against an angle of 30, then AB = 2 * AH = 2 * 1 = 2 cm.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 4 – 1 = 3.

BH = √3 cm.

Determine the perimeter of the trapezoid.

R avsd = BC + AD + 2 * AC = 3 + 5 + 2 * 2 = 12 cm.

Determine the area of the trapezoid.

Savsd = (ВС + АD) * ВН / 2 = (3 + 5) * √3 / 2 = 4 * √3 cm2.

Answer: The perimeter is 12 cm, the area is 4 * √3 cm2.