AMC 8 · 2017 · #18

Grade 8 geometry-2d

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

pythagorean-theoremarea-triangles area-differenceidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A non-convex quadrilateral ABCD has a right angle at C (so ∠ BCD = 90°) with legs BC = 4 and CD = 3. The other two sides are AB = 12 and AD = 13. The picture shows that vertex C pokes inward, so the quadrilateral looks like the big triangle △ ABD with the small triangle △ BCD scooped out. Find the area of ABCD.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The shape is an awkward non-convex quadrilateral, but the diagonal BD splits it into two right triangles we already understand. Tool #7 (Identify Subproblems) breaks the area question into three pieces: (a) find area of △ BCD, (b) find BD so we can study △ ABD, (c) find area of △ ABD, then subtract. Tool #1 (Draw a Diagram) is the natural companion — sketching the figure and drawing diagonal BD makes the 3-4-5 and 5-12-13 right triangles jump out, which is the whole shortcut. Tool #3 (Eliminate Possibilities) gives a quick sanity check at the end against the five answer choices.

1STEP 1

Draw diagonal BD: it splits ABCD into inner △ BCD and outer △ ABD, so area(ABCD) = area(△ ABD) - area(△ BCD).

Area(ABCD) = Area(△ ABD) - Area(△ BCD)

2STEP 2

The right angle at C makes legs BC and CD the base and height, so area(△ BCD) = ½·4·3 = 6.

Area(△ BCD) =

\frac{1}{2}

· BC · CD =

\frac{1}{2}

· 4 · 3 = 6

3STEP 3

Pythagoras in △ BCD gives the diagonal BD = √(3²+4²) = 5 — the classic 3-4-5 triple.

BD = √(BC² + CD²) = √(4² + 3²) = √(25) = 5

4STEP 4

Test △ ABD (sides 12, 5, 13): since 12² + 5² = 169 = 13², the converse of Pythagoras makes ∠ ABD a right angle.

12² + 5² = 144 + 25 = 169 = 13² ✓

5STEP 5

With the right angle at B, legs AB = 12 and BD = 5 give area(△ ABD) = ½·12·5 = 30.

Area(△ ABD) =

\frac{1}{2}

· AB · BD =

\frac{1}{2}

· 12 · 5 = 30

6STEP 6

Subtract: area(ABCD) = 30 - 6 = 24, matching choice (B); the trap 30 forgets to remove the scooped-out piece.

Area(ABCD) = 30 - 6 = 24 → (B)

Answer

A quick reality check: △ ABD alone has area 30 and △ BCD has area 6, so the answer must sit strictly between 30 - 6 = 24 (carve out the dent) and 30 + 6 = 36 (if the dent were instead an outward bump). The non-convex picture forces the subtraction, giving 24. The trap answer (D) = 30 corresponds to forgetting that C is on the inside; (E) = 36 would be the convex case. So 24 is the only number consistent with both the arithmetic and the picture.

💡Key takeaway

This AMC 8 problem only needs Grade 8 Pythagorean theorem (and its converse) you already know — once you spot the 3-4-5 and 5-12-13 right triangles, it's just one subtraction!

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