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This problem trains for: AMC-10.

Two lines (L) and (P) intersect at M. Points A, M, B, C, D are equidistant in this order on (L) and points R, M, S, T are equidistant in this order on (P). If the area of the quadrilateral ARDT is equal to 160 square units, what is the area of the quadrilateral SBCT?

Let us denote the area of the triangle ARM by a. Then, the area of the triangle RMD is:

because it has the same height as the triangle ARM while the corresponding base is three times longer.

The area of the triangle AMT is:

since it has the same height as the triangle ARM but the corresponding base is twice as long.

Similarly, the area of the triangles: TMB , TBC and TCD is:

and the area of the quadrilateral ARDT is:

therefore we can solve for a:

The area of the quadrilateral SBCT is the sum:

Substituting for a: