AMC 8 · 2005 · #4
Grade 5 geometry-2dPick an answer.
AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 ÷ 4 step.
Subproblem 1: add the triangle's three sides to get its perimeter.
Adding decimals to the tenths place is Grade 5 arithmetic — line up the decimal points and add column by column.
5.NBT.B.7Identify SubproblemsSubproblem 2: the square shares that 24 cm perimeter, so divide by 4 for one equal 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.D.8Identify SubproblemsSubproblem 3: square that 6 cm side to get the area.
Area of a square is side × side, a Grade 3 standard. A 6-by-6 square holds 36 unit squares.
3.MD.C.7Identify SubproblemsWhen 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 ÷ 4 and side-squared are routine Grade 3 work.
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