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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?

figure

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

equation

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:

equation

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:

equation

and the area of the quadrilateral ARDT is:

equation

therefore we can solve for a:

equation

The area of the quadrilateral SBCT is the sum:

equation

Substituting for a:

equation