2001 AMC 12 Problem 22

Triangle $EFG$ has base $FG = \dfrac{1}{3}AB$ and height equal to the rectangle’s height, so its area is $\dfrac{1}{6}(70) = \dfrac{35}{3}.$ Because $EC \parallel AF,$ triangles $AFH$ and $CEH$ are similar with ratio $\dfrac{EC}{AF} = \dfrac{3}{2},$ so $\dfrac{EH}{EF} = \dfrac{3}{5}.$ Likewise $\dfrac{EJ}{EG} = \dfrac{3}{7}.$ Then $\dfrac{[EHJ]}{[EFG]}$ $= \dfrac{EH}{EF}\cdot\dfrac{EJ}{EG}$ $= \dfrac{3}{5}\cdot\dfrac{3}{7}$ $= \dfrac{9}{35},$ giving $$[EHJ] = \dfrac{9}{35}\cdot\dfrac{35}{3} = 3.$$ Thus, the correct answer is C.