In parallelogram ABCD, the angle CAD = 30 degrees, vertex B is removed from the diagonal AC by 2

In parallelogram ABCD, the angle CAD = 30 degrees, vertex B is removed from the diagonal AC by 2, and from the side AD by 7. Find the area of the parallelogram.

1. BH and BK are perpendiculars to the AB side and the AC diagonal, respectively.

2. The angles CAD and ACB are equal as internal angles lying crosswise. Angle АСВ = 30 °.

3. The BK leg is opposite an angle of 30 °. Therefore, the hypotenuse of the BC of the BSC triangle is twice the indicated leg:

BC = BK x 2 = 2 x 2 = 4 units.

4. Area of the parallelogram ABCD = BC x BH = 4 x 7 = 28 units ^ 2.

Answer: the area of the parallelogram ABCD is 28 units ^ 2.