AMC 8 · 2013 · #24
Grade 6 geometry-2dProblem
Squares ABCD, EFGH, and GHIJ are equal in area. Points C and D are the midpoints of sides IH and HE, respectively. What is the ratio of the area of the shaded pentagon AJICB to the sum of the areas of the three squares?
Pick an answer.
AMC 8 2013 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: Three squares of equal area are arranged in an L-shape: square $EFGH$ sits on the left, square $GHIJ$ on the right (sharing side $GH$), and a third square $ABCD$ sits on top of the line $EHI$ with its base $DC$ landing exactly on the midpoints of $HE$ and $IH$. Find the ratio of the shaded pentagon $AJICB$ to the total area of all three squares.
Givens: Three squares $ABCD$, $EFGH$, $GHIJ$ all have the same area; $D$ is the midpoint of $HE$ and $C$ is the midpoint of $IH$; The shaded region is pentagon $AJICB$ (vertices in order $A \to J \to I \to C \to B$); Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{7}{24}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{5}{12}$
Unknowns: The ratio $\dfrac{\text{area of pentagon } AJICB}{\text{area of the three squares combined}}$
Understand
Restated: Three squares of equal area are arranged in an L-shape: square $EFGH$ sits on the left, square $GHIJ$ on the right (sharing side $GH$), and a third square $ABCD$ sits on top of the line $EHI$ with its base $DC$ landing exactly on the midpoints of $HE$ and $IH$. Find the ratio of the shaded pentagon $AJICB$ to the total area of all three squares.
Givens: Three squares $ABCD$, $EFGH$, $GHIJ$ all have the same area; $D$ is the midpoint of $HE$ and $C$ is the midpoint of $IH$; The shaded region is pentagon $AJICB$ (vertices in order $A \to J \to I \to C \to B$); Answer choices: (A) $\tfrac{1}{4}$, (B) $\tfrac{7}{24}$, (C) $\tfrac{1}{3}$, (D) $\tfrac{3}{8}$, (E) $\tfrac{5}{12}$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem
The shaded pentagon is awkward — its diagonal side $AJ$ cuts across two squares. Tool #7 (Identify Subproblems) is the geometry workhorse: slice the pentagon along the horizontal line through $C$ and $I$ so it becomes a clean trapezoid on top plus a right triangle on the bottom. Tool #1 (Diagram) gives us coordinates so we can read lengths off the page instead of doing algebra; Tool #9 (Easier Problem) tells us to pick a friendly side length — set $s = 2$ — so every distance is a whole or half number and the ratio at the end still comes out the same.
Execute — Answer: C
5.G.A.2 Step 1 - Put the figure on a grid.
- Set side length $s = 2$ and place $G$ at the origin.
- Then square $EFGH$ has corners $G(0,0), H(0,2), E(-2,2), F(-2,0)$ and square $GHIJ$ has corners $G(0,0), H(0,2), I(2,2), J(2,0)$.
- The midpoints give $D = (-1, 2)$ and $C = (1, 2)$, so the top square $ABCD$ has $A(-1, 4)$ and $B(1, 4)$.
💡 Plotting named points on a coordinate grid to model a real figure is exactly the Grade 5 coordinate-plane standard.
3.MD.C.7 Step 2 - Shrink the question with Tool #9.
- Each square now has area $2 \times 2 = 4$, so the three squares together have area $12$.
- The ratio we want is $\dfrac{\text{pentagon area}}{12}$, so we only need to find the pentagon's area in this concrete case.
💡 Area of a square by side $\times$ side is the Grade 3 multiplication-as-area idea, used here to turn the abstract $s$ into the friendly number $4$.
6.G.A.3 Step 3 - Use Tool #7: cut the pentagon along the horizontal line $y = 2$ (the line through $D, H, C, I$).
- The slanted side $AJ$ goes from $A(-1, 4)$ down to $J(2, 0)$; halfway down ($y = 2$) it sits halfway across, at $x = \tfrac{-1 + 2}{2} = \tfrac{1}{2}$.
- Call that crossing point $K(\tfrac{1}{2}, 2)$.
- The pentagon splits into a top piece $A K C B$ and a bottom piece $K I J$.
💡 Finding where a segment between two known vertices meets a horizontal line is the Grade 6 "polygons in the coordinate plane" move.
6.G.A.1 Step 4 - Top piece $AKCB$ is a trapezoid sitting between $y = 2$ and $y = 4$.
- The top edge $AB$ has length $1 - (-1) = 2$; the bottom edge $KC$ has length $1 - \tfrac{1}{2} = \tfrac{1}{2}$; the height between them is $4 - 2 = 2$.
- Trapezoid area = (average of parallel sides) $\times$ height.
💡 Finding the area of a trapezoid by decomposition is exactly the Grade 6 "area of special quadrilaterals" standard.
6.G.A.1 Step 5 - Bottom piece $KIJ$ is a right triangle.
- Side $IJ$ is vertical with length $2$, and side $KI$ is horizontal with length $2 - \tfrac{1}{2} = \tfrac{3}{2}$, meeting at the right angle at $I$.
💡 Half-base-times-height for a right triangle on a coordinate grid is the same Grade 6 polygon-area tool.
6.RP.A.3 Step 6 - Add the two subareas to get the pentagon and form the requested ratio.
- The pentagon area equals $\tfrac{5}{2} + \tfrac{3}{2} = 4$, which exactly matches the area of one of the squares — a nice sanity check.
- The total square area is $12$.
💡 Comparing two areas with a single ratio that reduces to a unit fraction is straight Grade 6 ratio reasoning.
5.G.A.2 Put the figure on a grid. Set side length $s = 2$ and place $G$ at the origin. T 3.MD.C.7 Shrink the question with Tool #9. Each square now has area $2 \times 2 = 4$, so 6.G.A.3 Use Tool #7: cut the pentagon along the horizontal line $y = 2$ (the line throug 6.G.A.1 Top piece $AKCB$ is a trapezoid sitting between $y = 2$ and $y = 4$. The top edg 6.G.A.1 Bottom piece $KIJ$ is a right triangle. Side $IJ$ is vertical with length $2$, a 6.RP.A.3 Add the two subareas to get the pentagon and form the requested ratio. The penta Review
Reasonableness: The pentagon's area came out to $4$, which is exactly the area of one square. That feels right when you look at the picture: the pentagon overlaps the right square fully and steals a triangular slice from the right side of the top square, while leaving an equally sized triangular slice of the top square un-shaded on the left. The two missing/added triangles are congruent (both have legs $1$ and $\tfrac{3}{2}$ at $s=2$), so the trade is even and the pentagon ends up with exactly one square's worth of area. One square out of three squares is $\tfrac{1}{3}$, matching (C).
Alternative: Tool #16 (Change Focus / Cut-and-Paste) gives the slickest path. Extend $AJ$ to where it crosses the line $DC$, call the crossing point $K$. Triangles $\triangle ADK$ and $\triangle JIK$ are congruent (both right triangles with legs $s$ and $\tfrac{3s}{4}$), so removing $\triangle ADK$ from square $ABCD$ and adding $\triangle JIK$ to that region is a zero-net-area swap. The pentagon therefore has the same area as square $ABCD$ — namely $s^2$ — and the ratio is $\dfrac{s^2}{3s^2} = \dfrac{1}{3}$ without ever computing coordinates.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication and addition operations (Computing the area of one square as $2 \times 2 = 4$ and the combined three-square area as $3 \times 4 = 12$.)5.G.A.2Represent real-world and mathematical problems by graphing points (Placing the three squares on a coordinate grid with named vertices $A, B, C, D, E, H, I, J$ so distances can be read directly.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing pentagon $AJICB$ into trapezoid $AKCB$ (area $\tfrac{5}{2}$) and right triangle $KIJ$ (area $\tfrac{3}{2}$), then adding to get $4$.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Finding the crossing point $K(\tfrac{1}{2}, 2)$ where the pentagon's slanted side $AJ$ meets the horizontal line through $C$ and $I$.)6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Forming the final ratio $\dfrac{4}{12} = \dfrac{1}{3}$ between the pentagon and the three squares.)
⭐ This AMC 8 problem only needs Grade 6 polygon-area decomposition — split, add, compare — that you already know!
⭐ This AMC 8 problem only needs Grade 6 polygon-area decomposition — split, add, compare — that you already know!
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