In parallelogram ABCD, angle A = 30 ° AD = 16cm, M-middle BC

In parallelogram ABCD, angle A = 30 ° AD = 16cm, M-middle BC. AM intersects BD at point N, CN intersects AB at point P, AP = 6cm. Find the area of the parallelogram.

1. Consider triangles AND and BNM, they are similar (in two corners). From the similarity with a triangle, we write down the aspect ratio:
AD: BM = DN: BN = 16: 8 = 2: 1.
2. Consider triangles PNB and CND, they are also similar in the first attribute. Let’s write down the ratio of the parties:
BN: DN = BP: CD = 1: 2.
CD = AB (parallelogram sides), BP: AB = 1/2. It follows from this that P is the middle of AB, AB = 12 cm.
We find the area of the parallelogram through its sides and the sine of the angle between them:
S = a * b * sin α = AB * AD * sin 30 ° = 12 * 16 * 1/2 = 96 cm².
Answer: the area of the parallelogram is 96 cm².