lines

Let's draw a line straight down from  P  to the  x-axis  and call that point of intersection  S.

Now notice that the area of △PQR is equal to the area of △PQS minus the area of △PRS

Now since  P  is  6  units above the x-axis,

PS  =  6

From the picture we can see that PR has a steeper slope than PQ,

and so we can say the slope of PQ is  1  and the slope of PR is 3

Since  PQ  has a slope of  1,

rise / run  =  1

PS / QS  =  1

6 / QS  =  1

6  =  QS

QS  =  6

Since PR  has a slope of  3 ,

rise / run  =  3

PS / RS  =  3

6 / RS  =  3

6  =  3 * RS

2  =  RS

RS  =  2

Now we can find the area of  △PQS and the area of △PRS

area of △PQS   =   (1/2) * base * height   =   (1/2) * QS * PS   =   (1/2) * 6 * 6   =   18

area of △PRS   =   (1/2) * base * height   =   (1/2) * RS * PS   =   (1/2) * 2 * 6   =   6

And so...

area of △PQR   =   area of △PQS  -  area of △PRS   =   18 - 6   =   12

(Note... all areas are in square units)

Alternatively...we can do

area of  △PQR   =   (1/2) * base * height   =   (1/2) * QR * PS   =   (1/2) * 4 * 6   =   12