In Parallelogram ABCD, angle A = 60 degrees, diagonal BD is perpendicular to side AB.

In Parallelogram ABCD, angle A = 60 degrees, diagonal BD is perpendicular to side AB. The straight line passing through the middle of the segment BD – point M parallel to AD, intersects the side AB at point K, MK = 4 cm. Find the area of the parallelogram.

1. According to the problem statement, the straight line MK intersects BD in its middle and is parallel to AD.

Therefore, MK is the middle line of the triangle ABD.

2. The length of the middle line is equal to half of the side parallel to it.

That is, AD = MK x 2 = 4 x 2 = 8 cm.

3. We calculate the length of the side AB of a given parallelogram, which in a rectangular

triangle ABD, is a leg:

AB / AD = cosine ∠A. Cosine 60 ° = 1/2. AB = 8 x 1/2 = 4 cm.

4. We calculate the length of the leg BD of the triangle ABD:

BD / AD = sinus ∠A. Sine 60 ° = √3 / 2. ВD = 8 х √3 / 2 = 4√3 cm.

5. Calculate the area (S) of the triangle ABD:

S = AB x BD / 2 = 4 x 4√3 / 2 = 8√3 cm².

6. The area of ​​triangle ABD is equal to the area of ​​triangle BCD equal to it. Considering this,

calculate the area of ​​a given parallelogram:

8√3 x 2 = 16√3 cm².

Answer: the area of ​​a given parallelogram is 16√3 cm².