AMC 8 · 2014 · #6
Easy mode Grade 4Problem
Imagine 6 rectangles in a row. Every rectangle has the same width: 2.
But the rectangles have different lengths. Their lengths are 1,4,9,16,25, and 36.
Find the area of each rectangle, then add all 6 areas together.
What is the total area?
Pick an answer.
AMC 8 2014 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: Six rectangles each have width $2$. Their lengths are the first six perfect squares: $1, 4, 9, 16, 25, 36$. Find the sum of their areas.
Givens: There are $6$ rectangles; Every rectangle has the same width: $2$; Lengths are $1, 4, 9, 16, 25, 36$ (the first six perfect squares); Answer choices: (A) $91$, (B) $93$, (C) $162$, (D) $182$, (E) $202$
Unknowns: The total area of all six rectangles combined
Understand
Restated: Six rectangles each have width $2$. Their lengths are the first six perfect squares: $1, 4, 9, 16, 25, 36$. Find the sum of their areas.
Givens: There are $6$ rectangles; Every rectangle has the same width: $2$; Lengths are $1, 4, 9, 16, 25, 36$ (the first six perfect squares); Answer choices: (A) $91$, (B) $93$, (C) $162$, (D) $182$, (E) $202$
Plan
Primary tool: #4 Find a Pattern
Secondary: #7 Identify Subproblems
Each rectangle's area is $2 \times \text{length}$, so the total area is $2 \times 1 + 2 \times 4 + \dots + 2 \times 36$. The shared factor $2$ is a pattern (Tool #4) — every term has it — so we can factor it out and add the lengths just once instead of six times: $\text{total} = 2 \times (1 + 4 + 9 + 16 + 25 + 36)$. Tool #7 (Identify Subproblems) then splits the work into two clean pieces — first add the six numbers, then multiply by $2$.
Execute — Answer: D
3.MD.C.7 Step 1 - Write the total area as a sum of six rectangle areas.
- Each rectangle area is width $\times$ length $= 2 \times \ell$.
💡 Area $=$ width $\times$ length for a rectangle is the Grade 3 area standard.
3.OA.B.5 Step 2 - Notice the shared factor of $2$ in every term.
- Pull it out using the distributive property so we only have to add the lengths once.
💡 Factoring out a common factor is exactly the distributive property of multiplication over addition.
4.NBT.B.4 Step 3 - Add the six perfect squares inside the parentheses.
- Group them in easy pairs to keep the arithmetic clean: $(1 + 9) + (4 + 16) + (25 + 36) = 10 + 20 + 61$.
💡 Pairing numbers that add to round values (like $10$ and $20$) is a standard mental-math regrouping move.
4.NBT.B.5 Step 4 Multiply the sum by the common width $2$ to get the total area.
💡 A one-digit by two-digit multiplication is Grade 4 multi-digit arithmetic.
3.MD.C.7 Write the total area as a sum of six rectangle areas. Each rectangle area is wid 3.OA.B.5 Notice the shared factor of $2$ in every term. Pull it out using the distributiv 4.NBT.B.4 Add the six perfect squares inside the parentheses. Group them in easy pairs to 4.NBT.B.5 Multiply the sum by the common width $2$ to get the total area. Review
Reasonableness: Choice (A) $91$ is the sum of the lengths alone, before multiplying by the width $2$ — a tempting trap if you forget the last step. Doubling $91$ gives $182$, which is choice (D). Quick sanity check: the largest single rectangle is $2 \times 36 = 72$, and the six areas $2, 8, 18, 32, 50, 72$ are all well under $200$, so a total near $182$ is reasonable.
Alternative: Tool #1 (Make a List): compute each area separately — $2, 8, 18, 32, 50, 72$ — and add: $2 + 8 = 10$, $10 + 18 = 28$, $28 + 32 = 60$, $60 + 50 = 110$, $110 + 72 = 182$. Same answer (D), but with six multiplications instead of one.
CCSS standards used (min grade 4)
3.MD.C.7Relate area to the operations of multiplication and addition (Using area $=$ width $\times$ length for each rectangle to write the total as a sum of six products.)3.OA.B.5Apply properties of operations as strategies to multiply and divide (Factoring out the common width $2$ via the distributive property: $2\cdot1 + 2\cdot4 + \dots = 2 \times (1 + 4 + \dots)$.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Adding the six perfect squares $1 + 4 + 9 + 16 + 25 + 36 = 91$.)4.NBT.B.5Multiply a whole number of up to four digits by a one-digit whole number (Computing the final total $2 \times 91 = 182$.)
⭐ This AMC 8 problem only needs Grade 4 arithmetic — once you factor out the shared width, it's just one addition and one multiplication!
⭐ This AMC 8 problem only needs Grade 4 arithmetic — once you factor out the shared width, it's just one addition and one multiplication!
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