AMC 8 · 2014 · #14

Easy mode Grade 8

Problem

Picture a rectangle $ABCD$ . One side ( $AB$ ) has length $5$ and the next side ( $AD$ ) has length $6$ .

Right next to the rectangle, sharing the side $\overline{DC}$ , there is a right triangle $DCE$ . The right angle of the triangle is at corner $C$ . The two shapes together make a trapezoid, like in the picture.

The triangle and the rectangle have the same area.

Find the length of the slanted side $DE$ 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.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

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

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

#7 Identify Subproblems 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.

$$\text{Area}(ABCD) = 5 \times 6 = 30$$

💡 Area of a rectangle as length $\times$ width is a Grade 3 standard — the foundation everything else rests on.

#7 Identify Subproblems 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.

$$\tfrac{1}{2} \cdot DC \cdot CE = 30 \;\Rightarrow\; \tfrac{1}{2} \cdot 5 \cdot CE = 30 \;\Rightarrow\; CE = \dfrac{60}{5} = 12$$

💡 Plug in what you know, isolate the unknown — Grade 6 one-step equation solving.

#1 Draw a Diagram 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 $C$.

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

💡 A diagram check catches a wrong leg choice before it propagates — the cheapest insurance in geometry.

#6 Guess and Check 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 $\le 20$.

$$DE^2 = 5^2 + 12^2 = 25 + 144 = 169 \;\Rightarrow\; DE = \sqrt{169} = 13 \;\Rightarrow\; \textbf{(B)}$$

💡 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.

[1] #7 3.MD.C.7 Subproblem 1: find the rectangle's area. The two given side lengths are $5$ and

[2] #7 6.EE.B.7 Subproblem 2: find the missing leg $CE$. The triangle has the same area, $30$. I

[3] #1 3.MD.C.7 Sanity check by looking at the picture (Tool #1): the rectangle is $5$ tall and

[4] #6 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.7 Relate 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.7 Solve 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.7 Apply 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 $a^2 + b^2 = c^2$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 8)

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