In an isosceles trapezoid, the smaller base is 5 cm, height √ 3, find the perimeter if one of the corners = 150.

Let’s define the acute angles of the right-angled triangle ABН. Angle АBН = ABC – СBН = 150 – 90 = 60, angle BАН = 90 – 60 = 30.

The BH leg lies opposite the angle 30, then BH = AB / 2, AB = 2 * BH = 2 * √3 cm.

Since the trapezoid is isosceles, CD = AB = 2 * √3 cm.

AH ^ 2 = AB ^ 2 – BH ^ 2 = 12 – 3 = 9.

AH = 3 cm 11

Let’s draw the height of the CК. Triangles ABH and CDK are equal in hypotenuse and acute angle, then DK = AH = 3 cm.

НC = BC = 5 cm, since BCКН is a rectangle.

AD = AH + НK + DK = 3 + 5 + 3 = 11 cm.

Determine the perimeter of the trapezoid. Ravsd = (AB + BC + CD + AD) = 2 * √3 + 5 + 2 * √3 + 11 = 16 + 4 * √3 = 4 * (4 + √3) cm.

Answer: The perimeter of the trapezoid is 4 * (4 + √3) cm.