ABCD-parallelogram. EC = 3, AB = 7, AED = 15 degrees, B = 150 degrees. find the area ABCD.

1. Since ABCD is a parallelogram, then AD = BC, AB = CD = 7. CD consists of two segments DE and EC:
CD = DE + EC;
DE + 3 = 7;
DE = 7 – 3;
DE = 4.
2. In a parallelogram, the opposite angles are equal, then angle B = angle D = 150 degrees.
Consider triangle AED: angle AED = 15 degrees, angle EDA (angle D) = 150 degrees. By the theorem on the sum of the angles of a triangle:
angle AED + angle EDA + angle DAE = 180 degrees;
15 + 150 + angle DAE = 180;
angle DАЕ = 180 – 165;
angle DАЕ = 15 degrees.
Since the angle DАЕ = angle AED = 15 degrees, then the triangle AED is isosceles, AE is the base, AD = DE = 4 are the sides.
3. From point C we draw the height CH up to the intersection with the line containing the side AD.
Consider a triangle CHD: the angle CHD = 90 degrees (since CH is the height), CD = 7 is the hypotenuse (since it lies opposite the right angle), the angle CDH = 180 degrees is the angle EDA (since the angles CDH and EDA are adjacent).
angle CDH = 180 – 150 = 30 (degrees).
CH – leg, which lies opposite an angle of 30 degrees, therefore:
CH = CD / 2;
CH = 7/2;
CH = 3.5.
4. The area of ​​the parallelogram is:
S = ah,
where a is the side of the parallelogram, h is the height of the parallelogram drawn to the side a.
S = AD * CH;
S = 4 * 3.5 = 14.
Answer: S = 14.