AMC 8 · 2007 · #8
Grade 6 geometry-2dProblem
In trapezoid ABCD, AD is perpendicular to DC,
AD=AB=3, and DC=6. In addition, E is on DC, and BE is parallel to AD. Find the area of △BEC.
Pick an answer.
AMC 8 2007 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: Trapezoid $ABCD$ has $\overline{AD} \perp \overline{DC}$ with $AD = AB = 3$ and $DC = 6$. Point $E$ sits on $\overline{DC}$ so that $\overline{BE} \parallel \overline{AD}$. Find the area of $\triangle BEC$.
Givens: $\overline{AD} \perp \overline{DC}$, so $\angle ADC = 90^\circ$; $AD = AB = 3$; $DC = 6$; $E$ is on $\overline{DC}$ and $\overline{BE} \parallel \overline{AD}$; Answer choices: (A) $3$, (B) $4.5$, (C) $6$, (D) $9$, (E) $18$
Unknowns: The area of $\triangle BEC$
Understand
Restated: Trapezoid $ABCD$ has $\overline{AD} \perp \overline{DC}$ with $AD = AB = 3$ and $DC = 6$. Point $E$ sits on $\overline{DC}$ so that $\overline{BE} \parallel \overline{AD}$. Find the area of $\triangle BEC$.
Givens: $\overline{AD} \perp \overline{DC}$, so $\angle ADC = 90^\circ$; $AD = AB = 3$; $DC = 6$; $E$ is on $\overline{DC}$ and $\overline{BE} \parallel \overline{AD}$; Answer choices: (A) $3$, (B) $4.5$, (C) $6$, (D) $9$, (E) $18$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Tool #1 (Draw a Diagram) is the lead because the answer pops out as soon as we annotate the figure: with $\overline{AD} \perp \overline{DC}$ and $\overline{BE} \parallel \overline{AD}$, the piece $ABED$ has two pairs of parallel sides and a right angle, so it is a rectangle — and since $AD = AB = 3$, it is a $3 \times 3$ square. Tool #7 (Identify Subproblems) then splits the trapezoid into the labeled square $ABED$ plus the right triangle $BEC$, so the triangle's legs ($BE$ and $EC$) read straight off the picture. No algebra needed.
Execute — Answer: B
5.G.B.3 Step 1 - Annotate the figure.
- Mark $\angle ADC = 90^\circ$ at $D$ and the parallel marks on $\overline{AD}$ and $\overline{BE}$.
- Quadrilateral $ABED$ has $\overline{AB} \parallel \overline{DE}$ (both horizontal) and $\overline{AD} \parallel \overline{BE}$ (both vertical), so it is a parallelogram.
- The right angle at $D$ makes it a rectangle, and $AD = AB = 3$ makes it a square.
💡 A Grade 5 student classifies $ABED$ by its properties: two pairs of parallel sides plus a right angle plus equal adjacent sides forces "square."
4.MD.A.3 Step 2 - Read the triangle's legs from the diagram.
- Because $ABED$ is a square, $BE = 3$ and $DE = 3$.
- Point $E$ lies on $\overline{DC}$, so the rest of $\overline{DC}$ — segment $\overline{EC}$ — is what is left after taking out $DE$.
💡 Splitting the trapezoid into "square piece" $+$ "triangle piece" is the subproblem move. The triangle's base is whatever the square does not cover.
6.G.A.1 Step 3 - Triangle $BEC$ is right-angled at $E$ because $\overline{BE} \parallel \overline{AD} \perp \overline{DC}$.
- Its legs are $BE = 3$ and $EC = 3$, so apply the right-triangle area formula.
💡 Once the legs are known, the Grade 6 area formula closes the problem in one line.
5.G.B.3 Annotate the figure. Mark $\angle ADC = 90^\circ$ at $D$ and the parallel marks 4.MD.A.3 Read the triangle's legs from the diagram. Because $ABED$ is a square, $BE = 3$ 6.G.A.1 Triangle $BEC$ is right-angled at $E$ because $\overline{BE} \parallel \overline Review
Reasonableness: Cross-check by computing the trapezoid's area two ways. Direct formula: $[ABCD] = \tfrac{1}{2}(AB + DC) \cdot AD = \tfrac{1}{2}(3 + 6) \cdot 3 = \tfrac{27}{2} = 13.5$. Decomposition: $[\text{square } ABED] + [\triangle BEC] = 9 + 4.5 = 13.5$. The two totals match, confirming the triangle's area is $4.5$. Also, $\triangle BEC$ is only a corner of the trapezoid, so an answer larger than the whole trapezoid (like (E) $18$) would be impossible — $4.5$ sits comfortably inside.
Alternative: Tool #16 (Use Coordinates): Place $D$ at the origin with $\overline{DC}$ along the $x$-axis. Then $D = (0,0)$, $C = (6,0)$, $A = (0,3)$, $B = (3,3)$, and $E = (3,0)$. The triangle with vertices $B=(3,3)$, $E=(3,0)$, $C=(6,0)$ has horizontal leg $EC = 3$ and vertical leg $BE = 3$, so its area is $\tfrac{1}{2} \cdot 3 \cdot 3 = 4.5$. Same answer (B).
CCSS standards used (min grade 6)
5.G.B.3Understand that attributes belonging to a category of two-dimensional figures also belong to all subcategories of that category (Classifying $ABED$ as a square by checking parallel sides, a right angle, and equal adjacent sides.)4.MD.A.3Apply the area and perimeter formulas for rectangles, including finding an unknown side length (Using $DC = DE + EC$ on the segment $\overline{DC}$ to recover $EC = 6 - 3 = 3$.)6.G.A.1Find the area of right triangles and other triangles by composing and decomposing into rectangles and other shapes (Decomposing the trapezoid into a square plus a right triangle and applying the $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$ formula to $\triangle BEC$.)
⭐ When a trapezoid has one slanted side, drop a perpendicular from the top corner. The trapezoid splits into a rectangle (or square) plus a right triangle, and the triangle's legs read straight off the picture.
⭐ When a trapezoid has one slanted side, drop a perpendicular from the top corner. The trapezoid splits into a rectangle (or square) plus a right triangle, and the triangle's legs read straight off the picture.
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