AMC 10 · 2020 · #20

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglessimilar-trianglespythagorean-theoremcoordinate-geometry identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-trianglessimilar-triangles

📏 Long solution 💡 3 insights

Problem

Quadrilateral ABCD has two right angles: ∠ ABC = 90° and ∠ ACD = 90°. The diagonal AC has length 20, and CD = 30. The two diagonals AC and BD meet at point E, with AE = 5. Find the area of ABCD.

Pick an answer.

(A)

330

(B)

340

(C)

350

(D)

360

(E)

370

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): the diagonal AC cuts ABCD into two right triangles △ ACD and △ ABC. Compute each area separately and add. △ ACD is immediate since both legs are given. △ ABC needs the altitude from B to AC — call it BF, with F on AC. Tool #1 (Diagram): draw AC horizontally with A on the right, C at the origin, D above C, and mark E on AC. Since BF and CD are both perpendicular to AC, they are parallel, so △ EBF ∼ △ EDC by AA. Tool #13 (Algebra): let EF = x, use the similarity to express BF in terms of x, then apply the right-triangle altitude theorem BF² = AF · FC to get x.

1STEP 1

Easy half first: right triangle ACD has legs 20 and 30, so [ACD] = $\frac{1}{2}$ · 20 · 30 = 300.

[△ ACD] =

\frac{1}{2}

· 20 · 30 = 300

2STEP 2

For triangle ABC, take AC as base: [ABC] = 10 · BF, so we need BF. Both BF and CD ⊥ AC, so BF ∥ CD.

[△ ABC] =

\frac{1}{2}

· 20 · BF = 10 · BF, BF ∥ CD

3STEP 3

Around E: BF ∥ CD makes △ EBF ∼ △ EDC (AA), so $\frac{BF}{CD}$ = $\frac{EF}{EC}$ , i.e. $\frac{BF}{30}$ = $\frac{EF}{15}$ , giving BF = 2 · EF.

\frac{BF}{30}

\frac{EF}{15}

→ BF = 2 · EF

4STEP 4

Let EF = x, so BF = 2x. The altitude-on-hypotenuse rule BF² = AF · FC (AF = 5 - x, FC = x + 15) gives (2x)² = (5 - x)(x + 15).

(2x)² = (5 - x)(x + 15)

5STEP 5

Expanding gives x² + 2x - 15 = 0, so (x + 5)(x - 3) = 0; the positive root x = 3 yields BF = 6.

x² + 2x - 15 = 0 → x = 3, BF = 6

6STEP 6

Then [ABC] = 10 · BF = 10 · 6 = 60, and adding the halves [ABCD] = 300 + 60 = 360 — choice (D).

[ABCD] = 300 + 60 = 360 → (D)

Answer

360

Sanity-check the geometry: place C = (0, 0), A = (20, 0), D = (0, 30), E = (15, 0). The line BD goes through D and E with slope -2. Drop the foot of the altitude F = (18, 0); then B = (18, -6) satisfies BF = 6 and the right angle at B: BA = (2, 6), BC = (-18, 6) with dot product -36 + 36 = 0 ✓. Also |AB| = √(40), |BC| = √(360), and

\frac{1}{2}

√(40 · 360) =

\frac{1}{2}

√(14400) = 60 ✓. The total 300 + 60 = 360 matches choice (D), and the nearby distractors 330, 340, 350, 370 correspond to wrong BF values.

💡Key takeaway

This AMC 10 problem only needs Grade 8 similar-triangle and Pythagorean reasoning you already know — slice the quadrilateral with diagonal AC, get 300 from the easy right triangle, find altitude BF = 6 via similar triangles, then add 60. The answer is (D) 360.

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