AMC 10 · 2008 · #7
Grade 6 geometry-2dAn equilateral triangle of side length 10 is completely filled in by non-overlapping equilateral triangles of side length 1. How many small triangles are required?
Pick an answer.
AMC 10 2008 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 big equilateral triangle has side length $10$. It is packed completely, with no gaps and no overlaps, by little equilateral triangles that each have side length $1$. Count how many little triangles that takes.
Givens: The big triangle is equilateral with side length $10$; The filling pieces are equilateral triangles of side length $1$; The pieces do not overlap and leave no gaps — they tile the big triangle exactly; Answer choices: (A) $10$, (B) $25$, (C) $100$, (D) $250$, (E) $1000$
Unknowns: The number of unit triangles needed to fill the big triangle
Understand
Restated: A big equilateral triangle has side length $10$. It is packed completely, with no gaps and no overlaps, by little equilateral triangles that each have side length $1$. Count how many little triangles that takes.
Givens: The big triangle is equilateral with side length $10$; The filling pieces are equilateral triangles of side length $1$; The pieces do not overlap and leave no gaps — they tile the big triangle exactly; Answer choices: (A) $10$, (B) $25$, (C) $100$, (D) $250$, (E) $1000$
Plan
Primary tool: #5 Look for a Pattern
Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem
Side length $10$ is hard to picture all at once, so shrink the problem first (Tool #9): try side $1$, side $2$, side $3$ and actually draw each one (Tool #1). Counting those small cases gives $1, 4, 9$ — the perfect squares. That pattern (Tool #5) says a side-$n$ triangle needs $n^2$ pieces, and then side $10$ is just $10^2$. The pattern route is safer than juggling the area formula, because you can literally see and check each small case.
Execute — Answer: C
6.G.A.1 Step 1 Split a small triangle
- Start small.
- A side-$2$ equilateral triangle breaks into unit triangles by cutting each side at its midpoint and joining those marks: you get $3$ small triangles pointing up and $1$ pointing down, $4$ in all.
- Do the same with a side-$3$ triangle and it splits into $9$ unit triangles.
- Drawing them shows the pieces really do fill the shape with no gaps, some pointing up and some down.
💡 Cutting a shape into equal pieces you can see beats guessing at the whole thing.
4.OA.C.5 Step 2 Count row by row
- Look at one triangle by horizontal rows, from the bottom up.
- In a side-$n$ triangle the bottom row holds $n$ up-pointing pieces plus $n-1$ down-pointing pieces, which is $2n-1$ triangles.
- Each row above it loses one up and one down, so the counts drop by $2$: the rows read $2n-1,\ 2n-3,\ \dots,\ 3,\ 1$.
- For side $3$ that is $5+3+1=9$, matching the picture.
💡 Odd numbers piling up by $2$ is a rhythm you can predict without redrawing.
6.EE.A.1 Step 3 Spot the square
- List the totals as the side grows: side $1$ gives $1$, side $2$ gives $4$, side $3$ gives $9$, side $4$ gives $16$.
- These are exactly $1^2, 2^2, 3^2, 4^2$ — the perfect squares.
- The same fact shows up in the rows: adding the first $n$ odd numbers always makes $n^2$.
- So a side-$n$ triangle needs $n^2$ unit triangles.
💡 When a count list turns into square numbers, the rule is just side squared.
6.EE.A.1 Step 4 Apply it to side 10
- The big triangle has side $10$, so plug $n=10$ into the rule: the number of unit triangles is $10^2 = 10 \times 10 = 100$.
- That matches choice (C).
- A quick sanity note: the count grows like the square of the side, so scaling the side from $1$ up to $10$ multiplies the piece count by $10^2=100$, not by $10$ — which rules out the tempting answer $10$.
💡 Once you know the rule is side squared, the big case is one multiplication.
6.G.A.1 Start small. A side-$2$ equilateral triangle breaks into unit triangles by cutti 4.OA.C.5 Look at one triangle by horizontal rows, from the bottom up. In a side-$n$ trian 6.EE.A.1 List the totals as the side grows: side $1$ gives $1$, side $2$ gives $4$, side 6.EE.A.1 The big triangle has side $10$, so plug $n=10$ into the rule: the number of unit Review
Reasonableness: The answer $100$ passes an independent area check. Similar triangles scale area by the square of the length ratio, and the big triangle is $10$ times as long on each side as a unit triangle, so its area is $10^2=100$ times as large. Equal-area pieces that tile it perfectly must therefore number $100$. Both the row-by-row count and the area argument land on the same value, and $100$ is choice (C). The distractors are the classic traps: $10$ (forgetting to square), $250$ and $1000$ (over-scaling).
Alternative: Use area directly instead of counting. A unit equilateral triangle has area $\frac{\sqrt{3}}{4}\cdot 1^2$ and the big one has area $\frac{\sqrt{3}}{4}\cdot 10^2$. Dividing, the $\frac{\sqrt{3}}{4}$ cancels and leaves $\frac{10^2}{1^2}=100$, so $100$ unit triangles fit.
CCSS standards used (min grade 6)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Decomposing a small equilateral triangle into unit equilateral pieces to see how the tiling works.)4.OA.C.5Generate a number or shape pattern following a given rule (Counting each triangle row by row as the odd numbers $2n-1, 2n-3, \dots, 1$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Recognizing the counts as perfect squares $n^2$ and evaluating $10^2=100$ for the side-$10$ triangle.)
⭐ A triangle with side $n$ made of unit triangles needs $n \times n$ of them, so side $10$ takes $10^2 = 100$.
⭐ A triangle with side $n$ made of unit triangles needs $n \times n$ of them, so side $10$ takes $10^2 = 100$.
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