Find the area of the trapezoid in which the sides and the smaller base are 8 cm and an acute angle at the base of п / 3.

Let’s define an acute angle at the base in degrees.

Angle BAD = 180/3 = 60.

Let’s build the height of the HV.

In a right-angled triangle ABN Cos60 = AH / AB.

AH = AB * Cos60 = 8 * 1/2 = 4 cm.

By the Pythagorean theorem, BH ^ 2 = AB ^ 2 – AH ^ 2 = 64 – 16 = 48.

BH = 4 * √3 cm.

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

AH = (AD – BC) / 2.

AD = 2 * AH + BC = 2 * 4 + 8 = 18 cm.

Determine the area of the trapezoid.

Savsd = (ВС + АD) * ВН / 2 = (8 + 16) * 4 * √3 / 2 = 48 * √3 cm2.

Answer: The area of the trapezoid is 48 * √3 cm2.