∴ 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!