AMC 8 · 2005 · #4
Easy mode Grade 5Problem
Picture a square and a triangle. Their perimeters are the same.
The triangle has three sides: cm, cm, and cm.
What is the area of the square, in square centimeters?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A square and a triangle have equal perimeters. The triangle's three sides are $6.1$ cm, $8.2$ cm, and $9.7$ cm. Find the area of the square in square centimeters.
Givens: Triangle side lengths: $6.1$ cm, $8.2$ cm, $9.7$ cm; The square and triangle have equal perimeters; Answer choices: (A) $24$, (B) $25$, (C) $36$, (D) $48$, (E) $64$
Unknowns: The area of the square in cm$^2$
Understand
Restated: A square and a triangle have equal perimeters. The triangle's three sides are $6.1$ cm, $8.2$ cm, and $9.7$ cm. Find the area of the square in square centimeters.
Givens: Triangle side lengths: $6.1$ cm, $8.2$ cm, $9.7$ cm; The square and triangle have equal perimeters; Answer choices: (A) $24$, (B) $25$, (C) $36$, (D) $48$, (E) $64$
Plan
Primary tool: #7 Break into Subproblems
Secondary: #1 Draw a Diagram
The question chains three small tasks behind one short sentence, so Tool #7 (Break into Subproblems) keeps the work organized: (i) add the three triangle sides to get the shared perimeter, (ii) divide by $4$ to get the square's side length, (iii) square that side to get the area. Tool #1 (Draw a Diagram) is the supporting move — a quick sketch of the triangle and square reminds us that "equal perimeter" is the only link between the two shapes, and that the square's four equal sides give the $\div 4$ step.
Execute — Answer: C
5.NBT.B.7 Step 1 Subproblem 1: find the triangle's perimeter by adding its three sides.
💡 Adding decimals to the tenths place is Grade 5 arithmetic — line up the decimal points and add column by column.
3.MD.D.8 Step 2 - Subproblem 2: the square has the same perimeter, $24$ cm.
- Since all four sides of a square are equal, divide the perimeter by $4$ to get one side.
💡 The Grade 3 perimeter formula for a square is $P = 4s$, so $s = P/4$. Equal perimeters means the $24$ carries straight from the triangle to the square.
3.MD.C.7 Step 3 Subproblem 3: square the side to get the area.
💡 Area of a square is side $\times$ side, a Grade 3 standard. A $6$-by-$6$ square holds $36$ unit squares.
5.NBT.B.7 Subproblem 1: find the triangle's perimeter by adding its three sides. 3.MD.D.8 Subproblem 2: the square has the same perimeter, $24$ cm. Since all four sides o 3.MD.C.7 Subproblem 3: square the side to get the area. Review
Reasonableness: Cross-check the perimeter: $6.1 + 8.2 = 14.3$ and $14.3 + 9.7 = 24$, confirming $24$ cm. Then $24 \div 4 = 6$ and $6^2 = 36$, so (C) is consistent. The other choices fail a quick sanity test: (A) $24$ matches the perimeter, not the area; (E) $64 = 8^2$ would need a side of $8$, giving perimeter $32 \ne 24$; (D) $48$ isn't even a perfect square, so it can't be the area of a whole-number-sided square.
Alternative: Tool #1 (Draw a Diagram): sketch the triangle with sides labeled $6.1$, $8.2$, $9.7$ and walk around it counting cm — you arrive back at the start after $24$ cm. Draw a square next to it and walk around: $4$ equal sides totaling $24$ cm means each side is $6$ cm. The square visibly tiles into a $6 \times 6$ grid of unit squares, $36$ in all. Same answer (C).
CCSS standards used (min grade 5)
5.NBT.B.7Add, subtract, multiply, and divide decimals to hundredths (Adding the three triangle sides $6.1 + 8.2 + 9.7 = 24$ with decimals aligned to the tenths place.)3.MD.D.8Solve real-world and mathematical problems involving perimeters of polygons (Using $P = 4s$ for the square to recover side $= 24/4 = 6$ cm from the shared perimeter.)3.MD.C.7Relate area to the operations of multiplication and addition (Computing the area of the $6$-cm square as $6 \times 6 = 36$ cm$^2$.)
⭐ When two shapes share a perimeter, that one number is the only bridge between them — add the triangle's sides to cross the bridge, then the square's $\div 4$ and side-squared are routine Grade 3 work.
⭐ When two shapes share a perimeter, that one number is the only bridge between them — add the triangle's sides to cross the bridge, then the square's $\div 4$ and side-squared are routine Grade 3 work.
More like this
Same archetype — closest grade level first.
