AMC 8 · 2011 · #7
Grade 6 geometry-2dProblem
Each of the following four large congruent squares is subdivided into combinations of congruent triangles or rectangles and is partially bolded. What percent of the total area is partially bolded?
Pick an answer.
AMC 8 2011 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: Four congruent large squares are each cut into congruent triangles or rectangles, and some pieces in each square are drawn with bold (thick) outlines. Taking the four squares together as the whole, what percent of the combined area is enclosed by the bold outlines?
Givens: There are $4$ large congruent squares, so the combined area is $4$ times the area of one square; Top-left square: split into $4$ congruent vertical rectangles; $1$ of them is bolded; Top-right square: split into $4$ congruent small quadrants, and the top-right quadrant is cut by a diagonal into two equal triangles; $1$ triangle is bolded; Bottom-left square: split into $4$ quadrants; the bolded region is the bottom-left quadrant plus an adjacent right triangle sitting on top of it (legs equal to half the big side); Bottom-right square: split into $4$ quadrants; the bottom-left quadrant is bolded; Answer choices: (A) $12\tfrac{1}{2}$, (B) $20$, (C) $25$, (D) $33\tfrac{1}{3}$, (E) $37\tfrac{1}{2}$ (percent)
Unknowns: The combined bolded area, expressed as a percent of the combined area of all four large squares
Understand
Restated: Four congruent large squares are each cut into congruent triangles or rectangles, and some pieces in each square are drawn with bold (thick) outlines. Taking the four squares together as the whole, what percent of the combined area is enclosed by the bold outlines?
Givens: There are $4$ large congruent squares, so the combined area is $4$ times the area of one square; Top-left square: split into $4$ congruent vertical rectangles; $1$ of them is bolded; Top-right square: split into $4$ congruent small quadrants, and the top-right quadrant is cut by a diagonal into two equal triangles; $1$ triangle is bolded; Bottom-left square: split into $4$ quadrants; the bolded region is the bottom-left quadrant plus an adjacent right triangle sitting on top of it (legs equal to half the big side); Bottom-right square: split into $4$ quadrants; the bottom-left quadrant is bolded; Answer choices: (A) $12\tfrac{1}{2}$, (B) $20$, (C) $25$, (D) $33\tfrac{1}{3}$, (E) $37\tfrac{1}{2}$ (percent)
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
The whole problem is visual, so Tool #1 (Draw a Diagram) is the natural lead: redraw each square, mark its subdivisions, and shade the bolded piece so the fraction it covers becomes obvious. Tool #7 (Identify Subproblems) lets us handle one square at a time instead of fighting all four bolded regions at once. Tool #9 (Solve an Easier Related Problem) is the bookkeeping trick: set the side of one large square to $1$ so each big-square area is $1$ and the combined area is $4$ — every bolded piece is then a simple fraction, no variables needed.
Execute — Answer: C
3.MD.C.7 Step 1 - Set one large square to have side $1$, so its area is $1 \times 1 = 1$.
- The combined area of the four squares is $4 \times 1 = 4$.
- This is the Tool #9 move: replace the unknown side length with the easiest number that keeps the proportions.
💡 Multiplying side $\times$ side to get a square's area is the Grade 3 "area by multiplication" idea.
3.G.A.2 Step 2 - Top-left square: the diagram splits it into $4$ congruent vertical strips, and $1$ strip is bolded.
- Each strip is one of $4$ equal pieces of a unit square.
💡 Partitioning a shape into $4$ equal parts and naming one part as $\tfrac{1}{4}$ is the Grade 3 partition standard.
3.G.A.2 Step 3 - Top-right square: split into $4$ equal quadrants (each $\tfrac{1}{4}$ of the big square), and the top-right quadrant is cut by a diagonal into two equal right triangles.
- The bolded piece is one of those triangles, so it is half of $\tfrac{1}{4}$.
💡 First partition the big square into $4$ equal parts, then partition one of those parts into $2$ equal triangles — fractions of fractions.
6.G.A.1 Step 4 - Bottom-left square: redraw the bolded outline as two pieces glued together — the bottom-left quadrant (area $\tfrac{1}{4}$) and a right triangle sitting on its top edge.
- That triangle's legs each have length $\tfrac{1}{2}$ (half the big side), so its area is $\tfrac{1}{2}\times\tfrac{1}{2}\times\tfrac{1}{2}=\tfrac{1}{8}$.
- Add the two pieces.
💡 Tool #7: chopping the irregular bolded region into a square plus a triangle turns one hard area into two easy ones.
3.G.A.2 Step 5 - Bottom-right square: split into $4$ equal quadrants, and exactly $1$ quadrant (the bottom-left one) is bolded.
- That bolded piece is $\tfrac{1}{4}$ of the big square.
💡 One of $4$ equal pieces of a unit square has area $\tfrac{1}{4}$ — a direct read from the partition.
5.NF.A.1 Step 6 - Add the four bolded pieces.
- Rewrite everything with denominator $8$ so the unlike fractions can be combined cleanly.
💡 The four subproblems collapse into one sum once they share the common denominator $8$ (Grade 5 unlike-denominator addition).
6.RP.A.3 Step 7 Compare the total bolded area $1$ to the combined area $4$, then convert that ratio to a percent.
💡 Turning a part-of-whole fraction into a percent is Grade 6 ratio reasoning.
3.MD.C.7 Set one large square to have side $1$, so its area is $1 \times 1 = 1$. The comb 3.G.A.2 Top-left square: the diagram splits it into $4$ congruent vertical strips, and $ 3.G.A.2 Top-right square: split into $4$ equal quadrants (each $\tfrac{1}{4}$ of the big 6.G.A.1 Bottom-left square: redraw the bolded outline as two pieces glued together — the 3.G.A.2 Bottom-right square: split into $4$ equal quadrants, and exactly $1$ quadrant (t 5.NF.A.1 Add the four bolded pieces. Rewrite everything with denominator $8$ so the unlik 6.RP.A.3 Compare the total bolded area $1$ to the combined area $4$, then convert that ra Review
Reasonableness: The four bolded pieces are $\tfrac{1}{4}$, $\tfrac{1}{8}$, $\tfrac{3}{8}$, $\tfrac{1}{4}$ of one big square — none larger than half — so their sum should sit a little above $1$ big square. It comes out to exactly $1$, which divided by the $4$ big squares gives $\tfrac{1}{4}=25\%$. That value sits between choices (B) $20\%$ and (D) $33\tfrac{1}{3}\%$, exactly where a "slightly more than a quarter of each" picture should land.
Alternative: Tool #16 (Count the Complement): in each square, find the unbolded fraction and subtract. Top-left: unbolded $=\tfrac{3}{4}$. Top-right: unbolded $=\tfrac{7}{8}$. Bottom-left: unbolded $=\tfrac{5}{8}$. Bottom-right: unbolded $=\tfrac{3}{4}$. Total unbolded $= \tfrac{6+7+5+6}{8}=\tfrac{24}{8}=3$. Total bolded $= 4-3 = 1$, and $\tfrac{1}{4}=25\%$. Same answer, different bookkeeping.
CCSS standards used (min grade 6)
3.MD.C.7Relate area to multiplication and to addition (Setting the area of one large square to side $\times$ side $= 1 \times 1 = 1$ and the combined area to $4 \times 1 = 4$.)3.G.A.2Partition shapes into parts with equal areas; express each part as a unit fraction (Reading off $\tfrac{1}{4}$, $\tfrac{1}{8}$, and $\tfrac{1}{4}$ directly from the partitioned top-left, top-right, and bottom-right squares.)6.G.A.1Find the area of right triangles by composing into rectangles (Computing the bottom-left bolded triangle's small piece as $\tfrac{1}{2}\times\tfrac{1}{2}\times\tfrac{1}{2}=\tfrac{1}{8}$ alongside the quadrant.)5.NF.A.1Add and subtract fractions with unlike denominators (Combining $\tfrac{1}{4}+\tfrac{1}{8}+\tfrac{3}{8}+\tfrac{1}{4}=1$ with a common denominator of $8$.)6.RP.A.3Use ratio and rate reasoning, including percents (Converting the part-to-whole ratio $\tfrac{1}{4}$ into the percent $25\%$.)
⭐ Set each big square's area to $1$, read each bolded piece as a fraction, then add — the four pieces sum to $1$ out of $4$, which is $25\%$.
⭐ Set each big square's area to $1$, read each bolded piece as a fraction, then add — the four pieces sum to $1$ out of $4$, which is $25\%$.
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