AMC 8 · 2005 · #13

Easy mode Grade 3
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Problem

Picture a six-sided shape with corners labeled A,B,C,D,E,FA, B, C, D, E, F. Every corner is a right angle, so the shape is made of straight horizontal and vertical sides only. (The figure looks like a big rectangle with a smaller rectangle cut out of one corner.)

The whole shape has area 5252. Three of the side lengths are given:

  • AB=8AB = 8
  • BC=9BC = 9
  • FA=5FA = 5

The other two sides are DEDE and EFEF. What is DE+EFDE + EF?

Pick an answer.

(A)
7
(B)
8
(C)
9
(D)
10
(E)
11
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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

#1 Draw a Diagram 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.
$$\text{Rectangle } ABCO: \text{ width } = AB = 8,\;\; \text{height} = BC = 9$$

💡 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.

#7 Identify Subproblems 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 $FA + DE = BC$.
$$FA + DE = BC \;\Rightarrow\; 5 + DE = 9 \;\Rightarrow\; DE = 4$$

💡 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."

#7 Identify Subproblems 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 $EF$.
$$52 = 8 \times 9 \;-\; EF \times DE \;=\; 72 - 4\,EF \;\Rightarrow\; 4\,EF = 20 \;\Rightarrow\; EF = 5$$

💡 "Big rectangle minus small rectangle" is the cleanest way to handle any L-shape area.

#7 Identify Subproblems 3.OA.D.8 Step 4

Add the two pieces to answer the question.

$$DE + EF = 4 + 5 = 9 \;\Rightarrow\; \textbf{(C)}$$

💡 Combine the two subproblem answers — the last step of any split-it-up plan.

[1] #1 3.G.A.1 Draw the picture and complete the rectangle. Sketch $ABCDEF$ as an L with the la
[2] #7 3.MD.D.8 Subproblem 1: find $DE$ from the vertical sides. On the left of the figure, the
[3] #7 3.MD.C.7 Subproblem 2: find $EF$ from the area. The L's area is the big rectangle minus t
[4] #7 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.1 Understand 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.7 Relate 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.8 Solve 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.8 Solve 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.