Consider ΔAQD. This triangle has the area
[tex]A_{AQD}=\dfrac{1}{2}\cdot AD\cdot H,[/tex]
where H is the heigh drawn from the point Q to the side AD.
Note that the height H is also the height of the parallelogram. So the area of the parallelogram ABCD is
[tex]A_{ABCD}=AD\cdot H.[/tex]
From these two statements you can conclude that
[tex]A_{ABCD}=2A_{AQD}.[/tex]
Now consider ΔDMQ. The ratio between the area of triangles DMQ and AQD is
[tex]\dfrac{A_{\triangle DMQ}}{A_{\triangle AQD}}=\dfrac{\frac{1}{2}\cdot MQ\cdot h}{\frac{1}{2}\cdot AQ\cdot h}=\dfrac{MQ}{AQ}.[/tex]
Since [tex]2AM=3QM,[/tex] you have that
[tex]\dfrac{QM}{AQ}=\dfrac{QM}{QM+AM}=\dfrac{QM}{QM+\frac{3}{2}QM}=\dfrac{2}{5}[/tex]
and
[tex]\dfrac{A_{\triangle DMQ}}{A_{\triangle AQD}}=\dfrac{2}{5}.[/tex]
Thus,
[tex]\dfrac{A_{\triangle DMQ}}{A_{ABCD}}=\dfrac{A_{\triangle DMQ}}{2A_{\triangle AQD}}=\dfrac{1}{2}\cdot \dfrac{2}{5}=\dfrac{1}{5}.[/tex]
Answer: correct choice is B
