AMC 8 · 2007 · #23
Grade 6 geometry-2dProblem
What is the area of the shaded pinwheel shown in the 5×5 grid?
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: A $5 \times 5$ unit grid contains a shaded pinwheel figure. The pinwheel's four blades meet at the center of the grid, and the figure's outline is drawn on grid lines and lines through the center. Find the area of the shaded pinwheel, in square units.
Givens: The grid is a $5 \times 5$ square, made of $25$ unit squares; The pinwheel has four-fold rotational symmetry about the center of the grid; The center of the grid is at $(2.5, 2.5)$ when the bottom-left corner is the origin; Each blade's outer edge meets the boundary of the grid; the unshaded region is what remains inside the $5 \times 5$ square; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $10$, (E) $12$
Unknowns: The area of the shaded pinwheel, in square units
Understand
Restated: A $5 \times 5$ unit grid contains a shaded pinwheel figure. The pinwheel's four blades meet at the center of the grid, and the figure's outline is drawn on grid lines and lines through the center. Find the area of the shaded pinwheel, in square units.
Givens: The grid is a $5 \times 5$ square, made of $25$ unit squares; The pinwheel has four-fold rotational symmetry about the center of the grid; The center of the grid is at $(2.5, 2.5)$ when the bottom-left corner is the origin; Each blade's outer edge meets the boundary of the grid; the unshaded region is what remains inside the $5 \times 5$ square; Answer choices: (A) $4$, (B) $6$, (C) $8$, (D) $10$, (E) $12$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The pinwheel has slanted edges and a complicated outline, but the unshaded region is much friendlier — it's just four corner unit squares and four side triangles. That is the signature of Tool #16 (Count the Complement): measure what is easy and subtract from the whole. Tool #1 (Draw a Diagram) sets the bottom-left corner at $(0,0)$ so the center sits at $(2.5, 2.5)$ and every blade tip has clean coordinates. Tool #7 (Identify Subproblems) breaks the unshaded region into two simple pieces — four $1 \times 1$ squares and four congruent triangles — each of which falls to a basic Grade 6 area formula.
Execute — Answer: B
5.G.A.1 Step 1 - Set up coordinates.
- Place the bottom-left corner of the grid at $(0, 0)$ so the grid covers $0 \le x \le 5$ and $0 \le y \le 5$.
- The center of the grid is at $(2.5, 2.5)$, and the total area is $5 \times 5 = 25$.
- We will compute the unshaded area and subtract from $25$.
💡 Putting the figure on a coordinate grid is a Grade 5 graphing move — it gives every vertex a clean number.
3.MD.C.7 Step 2 - Count the four unshaded corner squares.
- At each corner of the grid the pinwheel leaves a $1 \times 1$ square uncovered (for example, the bottom-left corner $(0,0)$ to $(1,1)$).
- Each has area $1$.
💡 Four identical unit squares — a Grade 3 area-of-a-square sum.
6.G.A.1 Step 3 - Find the area of one of the four unshaded side triangles.
- Look at the bottom triangle: its vertices are $(1, 0)$, $(4, 0)$, and the center $(2.5, 2.5)$.
- Take $(1,0)$ to $(4,0)$ as the base — it lies on the $x$-axis, so its length is $4 - 1 = 3$.
- The height is the perpendicular distance from $(2.5, 2.5)$ down to the $x$-axis, which is $2.5$.
💡 When the base sits on a grid line, the height is just the vertical distance from the far vertex — a Grade 6 triangle-area setup.
6.G.A.1 Step 4 - By the four-fold symmetry of the pinwheel, the side triangles on the top, left, and right are congruent to the bottom one.
- So the four side triangles together have area $4 \times 3.75 = 15$.
💡 Rotating the bottom triangle a quarter turn at a time produces the other three, so all four have the same area.
6.G.A.1 Step 5 Subtract the total unshaded area from the area of the whole grid to get the shaded pinwheel area.
💡 Total minus the easy-to-measure complement is the standard "count what's left" move.
5.G.A.1 Set up coordinates. Place the bottom-left corner of the grid at $(0, 0)$ so the 3.MD.C.7 Count the four unshaded corner squares. At each corner of the grid the pinwheel 6.G.A.1 Find the area of one of the four unshaded side triangles. Look at the bottom tri 6.G.A.1 By the four-fold symmetry of the pinwheel, the side triangles on the top, left, 6.G.A.1 Subtract the total unshaded area from the area of the whole grid to get the shad Review
Reasonableness: The shaded pinwheel clearly fits inside the $5 \times 5$ grid, so its area must be less than $25$. Picture each blade as roughly the size of a $1 \times 1.5$ strip narrowed near the center — four blades that small total a few square units, so something around $6$ feels right. Answer $6$ also gives a clean split: shaded $6$, unshaded $19$, sum $25$. The other answer choices give awkward complements ($4 \to 21$, $8 \to 17$, $10 \to 15$, $12 \to 13$); none break into a simple "four squares $+$ four congruent triangles" picture the way $6$ does.
Alternative: Use the pinwheel's symmetry directly. The shaded region is built from four congruent blades, and each blade splits along the line through the center into two congruent triangles, giving $8$ congruent small triangles in all. Take the one with vertices $(2.5, 2.5)$, $(0, 4)$, $(1, 4)$: the base from $(0,4)$ to $(1,4)$ has length $1$, and the perpendicular distance from $(2.5, 2.5)$ to the line $y = 4$ is $4 - 2.5 = 1.5$. So one small triangle has area $\tfrac{1}{2} \times 1 \times 1.5 = 0.75$, and $8 \times 0.75 = 6$. Same answer (B).
CCSS standards used (min grade 6)
5.G.A.1Use a pair of perpendicular number lines (axes) to define a coordinate system (Placing the grid on a coordinate plane with the bottom-left corner at $(0,0)$ so the center is at $(2.5, 2.5)$ and every blade tip has clean coordinates.)3.MD.C.7Relate area to the operations of multiplication; find areas of rectangles by tiling and by side-length multiplication (Computing the area of each $1 \times 1$ corner square as $1$ and summing four of them to get $4$.)6.G.A.1Find the area of triangles, quadrilaterals, and polygons by composing into rectangles or decomposing into triangles (Using $\tfrac{1}{2} \times \text{base} \times \text{height}$ on each side triangle (base $3$, height $2.5$) and subtracting the unshaded total from $25$.)
⭐ The pinwheel itself has slanted edges, but the empty space around it is plain: four unit squares and four equal triangles. Add those up, subtract from $25$, and the pinwheel's area falls out as $6$.
⭐ The pinwheel itself has slanted edges, but the empty space around it is plain: four unit squares and four equal triangles. Add those up, subtract from $25$, and the pinwheel's area falls out as $6$.
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