In the rectangular coordinate system above, the area of triangle PQR

In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3

The area of a triangle equals to \(area=\frac{1}{2}*base*height\).

Now the base of triangle PQR (or simply the length of line segment PQ) is 4 units (the difference between x-coordinates of the points P and Q), and the height is also 4 units (the altitude from the point Q to the x-axis equals to 4 units). So \(area_{PQR}=\frac{1}{2}*4*4=8\);

Similarly the base of triangle LMN is 12 units (the difference between x-coordinates of the points L and N), and the height is also 12 units (the altitude from the point M to the x-axis equals to 12 units). So \(area_{LMN}=\frac{1}{2}*12*12=72\);

Thus \(\frac{area_{PQR}}{area_{LMN}}=\frac{8}{72}=\frac{1}{9}\).

Answer: A.

As for the difficulty level, I'd say around 600.

Hope it's clear.

In the rectangular coordinate system above the area of triangle PQR is what fraction of the area triangle LMN?

A. 1/9
B. 1/8
C. 1/6
D. 1/5
E. 1/3