In an isosceles trapezoid ABCD, the diagonal AC is perpendicular to the side of CD

In an isosceles trapezoid ABCD, the diagonal AC is perpendicular to the side of CD. Find the area of the trapezoid if CAD = 30 degrees, AD = 12 cm.

The AСD triangle is rectangular, the angle СAD = 30 °, so the angle СDA = 90 – 30 = 60 °.

Since the trapezoid is isosceles, the angle ВAD = 60 °, and the side AB = СD = 12: 2 = 6 cm (the ASD triangle is rectangular with an angle of 30 °, the leg lying opposite the angle of 30 ° is equal to half the hypotenuse. That is, 12: 2 = 6 cm)

The ABC triangle is isosceles, since the BCA angle = the СBP angle = 30 ° crosswise with parallel BC and ABP, the angle BAC = 30 °. Since the triangle ABC is isosceles, then AB = BC = 6cm. The area is (6 + 12): 2 * 3√3 = 27√3 cm².

The height of this trapezoid is 3√3, is found from the triangle ABН by the Pythagorean theorem 6² – 3² = 27.

The answer is 27√3 cm².