math, geometry
posted by Anonymous .
The bases of trapezoid ABCD are \overline{AB} and \overline{CD}. Let P be the intersection of diagonals \overline{AC} and \overline{BD}. If the areas of triangles ABP and CDP are 8 and 18, respectively, then find the area of trapezoid ABCD.

Let x=AB and y=CD
Triangle PAB is similar to PCD so the area of PAB over the area of PCD is x^2/y^2
Plugging in 8 and 18 gives us x/y=2/3
So AP/PC=AB/CD=2/3. Triangle ABP and BCP have the same height to AC so...
The area of ADP/ABP=3/2
So the area of ADP=3/2*8=12
That means the area of the trapezoid is 8+18+12+12=50