AMC 8 · 2014 · #14
Grade 8 geometry-2dProblem
Rectangle ABCD and right triangle DCE have the same area. They are joined to form a trapezoid, as shown. What is DE?
Pick an answer.
AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Rectangle side $AB = 5$, so $DC = 5$; Rectangle side $AD = 6$, so $BC = 6$; $\triangle DCE$ has a right angle at $C$, with legs $DC$ and $CE$; $\text{Area}(\triangle DCE) = \text{Area}(ABCD)$; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Unknowns: The length $DE$ (the hypotenuse of $\triangle DCE$)
Understand
Restated: 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.
Givens: Rectangle side $AB = 5$, so $DC = 5$; Rectangle side $AD = 6$, so $BC = 6$; $\triangle DCE$ has a right angle at $C$, with legs $DC$ and $CE$; $\text{Area}(\triangle DCE) = \text{Area}(ABCD)$; Answer choices: (A) $12$, (B) $13$, (C) $14$, (D) $15$, (E) $16$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #6 Guess and Check
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 $13$.
Execute — Answer: B
3.MD.C.7 Step 1 - Subproblem 1: find the rectangle's area.
- The two given side lengths are $5$ and $6$, so multiply them.
💡 Area of a rectangle as length $\times$ width is a Grade 3 standard — the foundation everything else rests on.
6.EE.B.7 Step 2 - Subproblem 2: find the missing leg $CE$.
- The triangle has the same area, $30$.
- Its height is the shared side $DC = 5$, and its base is the unknown leg $CE$.
- Set up the triangle area formula and solve.
💡 Plug in what you know, isolate the unknown — Grade 6 one-step equation solving.
3.MD.C.7 Step 3 - Sanity check by looking at the picture (Tool #1): the rectangle is $5$ tall and $6$ wide.
- The triangle leg $CE$ is on the same line as $BC$ but $12$ long — exactly twice as wide as the rectangle.
- That matches the figure, which shows $E$ stretched well past $C$.
💡 A diagram check catches a wrong leg choice before it propagates — the cheapest insurance in geometry.
8.G.B.7 Step 4 - Subproblem 3: find the hypotenuse $DE$.
- The right triangle has legs $5$ and $12$.
- Either apply the Pythagorean theorem, or recognize the Pythagorean triple $(5, 12, 13)$ directly — that is the Tool #6 (Guess and Check) shortcut for legs $\le 20$.
💡 Pythagorean theorem is a Grade 8 standard; memorizing the small triples $(3,4,5)$, $(5,12,13)$, $(8,15,17)$ turns it into instant recall.
3.MD.C.7 Subproblem 1: find the rectangle's area. The two given side lengths are $5$ and 6.EE.B.7 Subproblem 2: find the missing leg $CE$. The triangle has the same area, $30$. I 3.MD.C.7 Sanity check by looking at the picture (Tool #1): the rectangle is $5$ tall and 8.G.B.7 Subproblem 3: find the hypotenuse $DE$. The right triangle has legs $5$ and $12$ Review
Reasonableness: 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: $\tfrac{1}{2} \cdot 5 \cdot 12 = 30 = 5 \cdot 6$. Both shapes have area $30$, as promised.
Alternative: Tool #3 (Eliminate Possibilities) on the choices: square each candidate hypotenuse and check whether the result minus $25$ (which is $DC^2$) is a perfect square that fits the area condition. (A) $12^2 - 25 = 119$ (no); (B) $13^2 - 25 = 144 = 12^2$ — gives $CE = 12$, area $= 30$ (yes); (C) $14^2 - 25 = 171$ (no); (D) $15^2 - 25 = 200$ (no); (E) $16^2 - 25 = 231$ (no). Only (B) works.
CCSS standards used (min grade 8)
3.MD.C.7Relate area to multiplication; find rectangle areas by tiling and by multiplying side lengths (Computing $\text{Area}(ABCD) = 5 \times 6 = 30$ and reading the diagram to confirm which side is shared with the triangle.)6.EE.B.7Solve one-step equations of the form $px = q$ (Solving $\tfrac{1}{2} \cdot 5 \cdot CE = 30$ for the missing leg $CE = 12$.)8.G.B.7Apply the Pythagorean theorem to find unknown side lengths in right triangles (Computing the hypotenuse $DE = \sqrt{5^2 + 12^2} = 13$, or recognizing the $(5, 12, 13)$ Pythagorean triple.)
⭐ 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 $a^2 + b^2 = c^2$.
⭐ 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 $a^2 + b^2 = c^2$.
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