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The bases of trapezoid ABCD are AB and CD. Let P be the intersection of diagonals AC and BD. If the areas of triangles ABP and
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Let ABCD be a parallelogram. Let M be the midpoint of AB and N be the midpoint of AD. Diagonal BD intersects CM and CN at P and
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Solution 1: Triangles ADE and BFE are similar since \overline{AD}\parallel\overline{BF}. Furthermore, the ratio of their areas is 9:1, so AE/BE = 3. Then CD = AB = AE + BE = 3BE + BE = 4BE. Triangles FEB and FDC are similar, so the area of triangle FDC is
DENSITY=MASS/VOLUME =24/4=7
