Find the area of a trapezoid, the bases of which have lengths 7 and 9

Find the area of a trapezoid, the bases of which have lengths 7 and 9, and one of the lateral sides is 15 and forms an angle of 150 ° with one of the bases

1. The sum of the one-sided angles A and B with parallel lines AD and BC is 180 °:

∠A + ∠B = 180 °, hence:
∠A = 180 ° – ∠B = 180 ° – 150 ° = 30 °.
2. Let’s draw the height of the trapezoid BH. In a right-angled triangle ABH, leg BH, opposite an angle of 30 °, is equal to half the hypotenuse:

AB = 15;
BH = 1/2 * AB = 1/2 * 15 = 15/2.
3. The area of the trapezoid is equal to the product of the height and the half-sum of the bases:

AD = 9;
BC = 6;
S = BH * (AD + BC) / 2;
S = 15/2 * (9 + 7) / 2 = 15/2 * 8 = 4 * 15 = 60.
Answer: 60.