AMC 8 · 2005 · #13
Grade 3 geometry-2dProblem
The area of polygon ABCDEF is 52 with AB=8, BC=9 and FA=5. What is DE+EF?
Pick an answer.
AMC 8 2005 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: Hexagon $ABCDEF$ is an L-shape with every corner a right angle. Going around in order, $AB = 8$ (top), $BC = 9$ (right side), and $FA = 5$ (left side). The polygon has area $52$. Find $DE + EF$.
Givens: $AB = 8$, $BC = 9$, $FA = 5$; Area of polygon $ABCDEF = 52$; Every interior angle of $ABCDEF$ is a right angle (an axis-aligned L-shape); Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Unknowns: The sum $DE + EF$
Understand
Restated: Hexagon $ABCDEF$ is an L-shape with every corner a right angle. Going around in order, $AB = 8$ (top), $BC = 9$ (right side), and $FA = 5$ (left side). The polygon has area $52$. Find $DE + EF$.
Givens: $AB = 8$, $BC = 9$, $FA = 5$; Area of polygon $ABCDEF = 52$; Every interior angle of $ABCDEF$ is a right angle (an axis-aligned L-shape); Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Tool #1 (Draw a Diagram) is the cracker: sketch the L-shape with the labels, then extend $AF$ down and $DC$ up until they meet at a new corner $O$. The L is now the big rectangle $ABCO$ with a smaller rectangle cut out of one corner — and that cut-out rectangle has sides $DE$ and $EF$. Tool #7 (Identify Subproblems) then breaks the question into two small pieces: (a) use the right-angle structure to find $DE$ from the vertical sides, and (b) use the area equation "big rectangle minus cut-out = $52$" to find $EF$. Choosing Tool #1 + Tool #7 over Tool #13 (Convert to Algebra) keeps the work at one short subtraction and one short equation — no system of equations needed.
Execute — Answer: C
3.G.A.1 Step 1 - Draw the picture and complete the rectangle.
- Sketch $ABCDEF$ as an L with the labeled sides, then extend $FA$ downward and $DC$ upward until they meet at a new corner $O$.
- Now $ABCO$ is a rectangle whose width is $AB = 8$ and whose height is $BC = 9$.
- The original L equals this rectangle with the small rectangle $FEDO$ cut out of one corner.
💡 Closing the L into a rectangle turns an unfamiliar 6-sided shape into a difference of two rectangles — shapes whose area is just length times width.
3.MD.D.8 Step 2 - Subproblem 1: find $DE$ from the vertical sides.
- On the left of the figure, the height is covered by $FA$ on top of the cut-out and $DE$ on the bottom of the cut-out — together they must match the full height $BC$ on the right.
- So $FA + DE = BC$.
💡 In any axis-aligned shape, the verticals on the left must sum to the same total as the verticals on the right — that is just "two paths between the same two horizontals have equal length."
3.MD.C.7 Step 3 - Subproblem 2: find $EF$ from the area.
- The L's area is the big rectangle minus the small cut-out rectangle $FEDO$, which has sides $DE = 4$ and $EF$.
💡 "Big rectangle minus small rectangle" is the cleanest way to handle any L-shape area.
3.OA.D.8 Step 4 Add the two pieces to answer the question.
💡 Combine the two subproblem answers — the last step of any split-it-up plan.
3.G.A.1 Draw the picture and complete the rectangle. Sketch $ABCDEF$ as an L with the la 3.MD.D.8 Subproblem 1: find $DE$ from the vertical sides. On the left of the figure, the 3.MD.C.7 Subproblem 2: find $EF$ from the area. The L's area is the big rectangle minus t 3.OA.D.8 Add the two pieces to answer the question. Review
Reasonableness: Plug the numbers back into the picture. With $DE = 4$ and $EF = 5$, the cut-out corner is a $4 \times 5$ rectangle of area $20$. The big rectangle has area $8 \times 9 = 72$. Subtracting, $72 - 20 = 52$, which matches the given area exactly. Magnitude check: $DE$ must be smaller than $BC = 9$ (it is only part of the left side), and $EF$ must be smaller than $AB = 8$ (it is only part of the top-to-bottom step), so $DE + EF < 9 + 8 = 17$ — answer $9$ comfortably fits. Choice (A) $7$ would force $4\,EF = 12$, giving area $72 - 12 = 60 \neq 52$; (E) $11$ would force $4\,EF = 28$, giving area $72 - 28 = 44 \neq 52$. Only $9$ works.
Alternative: Tool #1 (Draw a Diagram) again, but slice the L into two rectangles instead of subtracting one. Extend $EF$ to the right until it hits side $BC$ at point $P$. Now the L splits into the top rectangle $ABPF$ of size $8 \times FA = 8 \times 5 = 40$, and the bottom rectangle $PCDE$ of size $BC{-}FA \times CD = 4 \times CD$. Since the top width equals $AB = CD + EF$, $CD = 8 - EF$. Setting the total area to $52$: $40 + 4(8 - EF) = 52 \Rightarrow 32 - 4EF = 12 \Rightarrow EF = 5$, and $DE = BC - FA = 4$. Sum: $9$, choice (C).
CCSS standards used (min grade 3)
3.G.A.1Understand that shapes in different categories may share attributes, and that the shared attributes can define a larger category (Recognizing the L-shape's right-angle corners and rebuilding it as a big rectangle with a smaller rectangle cut out.)3.MD.C.7Relate area to the operations of multiplication and addition (including finding area by decomposing into non-overlapping rectangles) (Computing area of the L-shape as $\text{big rectangle area} - \text{cut-out rectangle area}$, i.e. $8 \times 9 - EF \times DE = 52$.)3.MD.D.8Solve real world and mathematical problems involving perimeters and side lengths of polygons (Using the right-angle structure $FA + DE = BC$ to solve for $DE = 9 - 5 = 4$.)3.OA.D.8Solve two-step word problems using the four operations (Combining $DE = 4$ and $EF = 5$ with addition to get the requested sum $DE + EF = 9$.)
⭐ Close the L-shape into a full rectangle, and the missing corner is a smaller rectangle. "Big rectangle minus cut-out = $52$" gives $EF$, and "left verticals add up to the right vertical" gives $DE$ — pure Grade 3 area arithmetic.
⭐ Close the L-shape into a full rectangle, and the missing corner is a smaller rectangle. "Big rectangle minus cut-out = $52$" gives $EF$, and "left verticals add up to the right vertical" gives $DE$ — pure Grade 3 area arithmetic.
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