Question

Point Q is a point on the side BC of a parallelogram ABCD. point M is on AQ so that 2AM=3QM. Ratio Area of DMQ : Area of ABCD is equal to:

(A) 1/6 (B) 1/5 (C) 1/10 (D) 2/5

Answer 1

Consider ΔAQD. This triangle has the area

`A_(AQD)=(1)/(2)\cdot AD\cdot H,`

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

`A_(ABCD)=AD\cdot H.`

From these two statements you can conclude that

`A_(ABCD)=2A_(AQD).`

Now consider ΔDMQ. The ratio between the area of triangles DMQ and AQD is

`(A_(\triangle DMQ))/(A_(\triangle AQD))=((1)/(2)\cdot MQ\cdot h)/((1)/(2)\cdot AQ\cdot h)=(MQ)/(AQ).`

Since
`2AM=3QM,` you have that

`(QM)/(AQ)=(QM)/(QM+AM)=(QM)/(QM+(3)/(2)QM)=(2)/(5)`

and

`(A_(\triangle DMQ))/(A_(\triangle AQD))=(2)/(5).`

Thus,

`(A_(\triangle DMQ))/(A_(ABCD))=(A_(\triangle DMQ))/(2A_(\triangle AQD))=(1)/(2)\cdot (2)/(5)=(1)/(5).`

Answer: correct choice is B