AMC 8 · 2014 · #14

Grade 8 geometry-2d

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

area-rectanglesarea-trianglespythagorean-theorem identify-subproblems ↑ Prerequisites: area-rectanglesarea-trianglespythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

Rectangle ABCD has sides AB = 5 and AD = 6. A right triangle DCE shares side DC with the rectangle (so DC = 5 is one leg) and has its right angle at C, with the other leg CE along the line BC extended. The rectangle and the triangle have the same area. Find the length of DE, the hypotenuse of the triangle.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The question hides three smaller questions inside one: (a) what is the rectangle's area? (b) given that area, how long is the missing leg CE? (c) given both legs, how long is the hypotenuse DE? That is Tool #7 (Identify Subproblems) at work — solve each piece, then chain them. Tool #1 (Draw a Diagram) helps us see that DC = 5 is the shared side, so it is the height of the triangle, with CE on the ground. Tool #6 (Guess and Check) lets us avoid heavy algebra on the last step: the legs come out to 5 and 12, and (5, 12, 13) is the Pythagorean triple every AMC student should recognize, so the hypotenuse must be 13.

1STEP 1

Subproblem 1 — the rectangle's area is just its two sides multiplied: 5 × 6 = 30.

Area(ABCD) = 5 × 6 = 30

2STEP 2

Subproblem 2 — both areas equal 30, and ½ · DC · CE = 30 with DC = 5 gives leg CE = 12.

\frac{1}{2}

· DC · CE = 30 →

\frac{1}{2}

· 5 · CE = 30 → CE =

\frac{60}{5}

= 12

3STEP 3

Diagram check (Tool #1): CE = 12 is exactly twice the rectangle's width 6, matching E stretched far past C in the figure.

CE = 12, while BC = 6, so E sits 12 - 6 = 6 beyond the rectangle.

4STEP 4

Subproblem 3 — legs 5 and 12 make DE the hypotenuse; a² + b² = c² gives DE = 13, the classic (5, 12, 13) triple.

DE² = 5² + 12² = 25 + 144 = 169 → DE = √(169) = 13 → (B)

Answer

The hypotenuse of a right triangle must be longer than either leg but shorter than their sum. Legs are 5 and 12, so DE must satisfy 12 ＜ DE ＜ 17. Only choices (B) 13, (C) 14, (D) 15, and (E) 16 survive — and of those, only 13 matches the well-known (5, 12, 13) triple. The area check also lines up:

\frac{1}{2}

· 5 · 12 = 30 = 5 · 6. Both shapes have area 30, as promised.

💡Key takeaway

This AMC 8 problem chains three quick steps — rectangle area, missing leg, then Pythagorean theorem — so the only Grade 8 idea you really need is a² + b² = c².

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