In an isosceles trapezoid, the smaller base is 5, and the height is the root of 3, find s if one angle is 150g.

Let in trapezium ABCD the height BH = √3 is given; <ABC = 150 °; BC = 5;

Find S (ABCD) = (AD + BC) / 2 * BH.

Find AD. Consider a triangle ABH, in which <BAD = 180 ° – <ABC = 180 – 150 ° = 30 °. In triangle ABH: AB = 2 * BH = 2 * √3, where the leg is known against <30 °.

We find: AH = (AB ^ 2 – BH ^ 2) = √ [(2 * √3) ^ 2 – (√3) ^ 2] = √ [(4 * 3 – 3)] = √ (12 – 3) = √9 = 3.

Then the lower base is on the upper base AD = BC + 2 * AH = 5 + 2 * 3 = 5 + 6 = 11.

S (ABCD) = (5 + 11) / 2 * √3 = 16/2 * √3 = 8 * √3.