∴ AB = DC - ①

M is the midpoint of CD.

∴ DM/DC = 1/2 - ②

From ① ②: DM/AB = DM/DC = 1/2.

∵ Quadrilateral ABCD is a parallelogram

∴△dmn∽△ class, similarity ratio DM/AB = 1/2.

∴△DMN and △△ half area ratio are the square of similarity ratio, that is, 1: 4.

If the area of △DMN is s, then the area of △ class is 4S.

In addition, from class △DMN∽△, we can also get: MN/AN = 1/2.

∴△AND covers an area of 2 square kilometers.

(Because the height of △ sum is equal to that of △DMN, their area ratio is the ratio of their bottoms, MN/AN = 1/2).

∴s△abd = s△ABN+s△ sum

= 4S+2S

= 6S

The area of parallelogram ABCD is 2×S△ABD = 2× 6S = 12S.

∴ area ratio of triangle DMN to parallelogram ABCD

S :( 12S)= 1 : 12

Good luck with your study!