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In the coordinate plane, triangle ABC has vertices A at (0, 0), B at (6, 0), and C at (0, 6). Point D lies on side AB at (4, 0), and point E lies on side AC at (0, 4). Segments CD and BE intersect at point P. What is the area of quadrilateral ADPE?
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Correct answer
D
Set up the two cevians. Segment CD runs from C (0, 6) to D (4, 0): slope is −3/2, so y = 6 − (3/2)x. Segment BE runs from B (6, 0) to E (0, 4): slope is −2/3, so y = 4 − (2/3)x.
Find P, their intersection. Set 6 − (3/2)x = 4 − (2/3)x. Then 2 = (3/2 − 2/3)x = (5/6)x, so x = 12/5 = 2.4, and y = 6 − (3/2)(2.4) = 2.4. So P = (2.4, 2.4).
Apply the shoelace formula to quadrilateral ADPE with vertices in order A (0, 0), D (4, 0), P (2.4, 2.4), E (0, 4). The area is ½ × |0(0 − 4) + 4(2.4 − 0) + 2.4(4 − 0) + 0(0 − 2.4)| = ½ × (9.6 + 9.6) = ½ × 19.2 = 9.6. The answer is D.
There is a faster path using symmetry. Reflecting across the line y = x swaps B with C and D with E, carrying segment CD onto BE, so P lies on y = x. Then x = 6 − (3/2)x gives (5/2)x = 6, so x = 2.4 with no system to solve, and the area is twice triangle ADP, which is 2 × (½ × 4 × 2.4) = 9.6.
A full check: the whole triangle has area 18, split by the cevians into ADPE = 9.6, DPB = 2.4, BPC = 3.6, and EPC = 2.4, summing to 18.
The distractors target wrong regions or steps. 3.6 (A) is triangle BPC, the opposite-corner region. 6 (B) misreads D and E as midpoints, turning the cevians into medians and P into the centroid (2, 2). 8 (C) computes triangle ADE, forgetting the boundary bends at P. 14.4 (E) subtracts only BPC from the whole triangle, forgetting the two side triangles DPB and EPC.
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