AMC 8 · 2003 · #21

Grade 8 geometry-2d

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

area-rectanglesarea-trianglespythagorean-theoremperimeter area-differenceidentify-subproblems ↑ Prerequisites: area-trianglesarea-rectanglespythagorean-theoremlinear-equations-one-var

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Trapezoid ABCD has area 164 cm², altitude 8 cm, leg AB = 10 cm, and leg CD = 17 cm. The parallel sides are the top BC and the bottom AD. Find the length of the shorter base BC.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Break into Subproblems

The trapezoid carries three independent facts (area, altitude, two slant sides), so Tool #7 (Break into Subproblems) splits the figure into pieces we already know how to handle: drop perpendiculars from B and C to AD, and the trapezoid becomes one rectangle plus two right triangles. Tool #1 (Draw a Diagram) makes the split visible — once the perpendiculars are drawn, the Pythagorean theorem cracks each side triangle. Tool #13 (Convert to Algebra) ties the pieces together: the area formula gives BC + AD = 41, and the triangle bases give AD = BC + 21. Solve the pair.

1STEP 1

Feed the area 164 and altitude 8 into A = ½(b₁+b₂)h; it collapses to one equation, BC + AD = 41.

164 =

\frac{1}{2}

(BC + AD) · 8 → 164 = 4(BC + AD) → BC + AD = 41

2STEP 2

Drop perpendiculars from B and C to AD: the trapezoid becomes a rectangle flanked by two right triangles, each with leg BE = CF = 8.

AD = AE + EF + FD, EF = BC, BE = CF = 8

3STEP 3

Pythagoras on each side triangle (legs paired with 8): 6-8-10 gives AE = 6, and 8-15-17 gives FD = 15.

AE = √(10² - 8²) = √(36) = 6, FD = √(17² - 8²) = √(225) = 15

4STEP 4

The long base is the two feet plus the rectangle's width EF = BC, so AD = BC + 21.

AD = AE + EF + FD = 6 + BC + 15 = BC + 21

5STEP 5

Substitute AD = BC + 21 into BC + AD = 41: 2·BC = 20, so BC = 10 — choice (B).

BC + (BC + 21) = 41 → 2BC = 20 → BC = 10 → (B)

Answer

Plug BC = 10 back into the picture. Then AD = 10 + 21 = 31, and the trapezoid area is

\frac{1}{2}

(10 + 31) · 8 =

\frac{1}{2}

· 41 · 8 = 164 cm², matching the given area exactly. The horizontal feet 6 and 15 combine with the altitude 8 to give hypotenuses √(6²+8²)=10 and √(15²+8²)=17, matching AB and CD. Every input is reproduced, so the answer is consistent. Among the choices, 10 is the only value that keeps the longer base AD = 31 positive and shorter than the trapezoid would need; values like 20 would force AD = 41 - 20 = 21, which is shorter than BC + 21 = 41 — a contradiction.

💡Key takeaway

A trapezoid with two slant sides is really a rectangle hiding between two right triangles. Drop the altitudes, use the Pythagorean theorem on each side triangle, then let the area formula seal the answer.

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