Consider a rectangle ABCD of area 90 units. The points P and Q trisect AB, and R bisects CD. The diagonal AC intersects the line segments PR and QR at M and N respectively, What is the area of the quadrilateral PQMN?

This problem can be solved in two ways and both are lengthy. You can use either coordinates or similar triangles. The below method explains the solution using coordinates.
Let A, B, C and D be (0,0), (3a,0), (3a,b) and (0,b) respectively => 3ab = 90 => ab = 30 (as area of rectangle is equal to 90)
So points P, Q and R will be (a,0), (2a,0) and (3a/2, b) respectively.
Now, let's find the points M and N by using the line equations of the diagonal AC, PR and QR.
Equation of AC is bx-3ay = 0, equation of PR is 2bx-ay=2ab and equation of QR is 2bx+ay=4ab.
Using these three equations we can find that the points M and N are (6a/5, 2b/5) and (12a/7, 4b/7) respectively.
Now, area of quadrilateral PQMN = area of triangle AQN - Area of triangle APM
=> Area of PQMN = $$\frac{1}{2}*2a*\frac{4b}{7}-\frac{1}{2}*a*\frac{2b}{5}$$ = $$\frac{13ab}{35}$$ = $$\frac{78}{7}$$ sq units