AMC 8 · 2016 · #2

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglesarea-rectangles identify-subproblemsarea-difference ↑ Prerequisites: area-rectanglesfraction-multiplication

📏 Short solution 💡 2 insights 📊 Diagram

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Problem

Rectangle ABCD has AB = 6 and AD = 8. Point M is the midpoint of side AD. Find the area of △ AMC.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

This is a 2D geometry problem with named points, so Tool #1 (Draw a Diagram) is the natural first move: sketch the rectangle, mark M as the midpoint of AD, and draw △ AMC. The picture immediately reveals a key fact — side AM lies along side AD of the rectangle, which is perpendicular to side DC. That means AM can serve as the base and DC gives the height for free. Tool #7 (Identify Subproblems) then splits the area calculation into two simple subproblems: (a) find the base AM, (b) find the perpendicular height, and (c) apply the triangle area formula.

1STEP 1

Sketch rectangle ABCD, mark M at the middle of side AD, and draw △ AMC — side AM sits on the rectangle's left edge.

Rectangle: AB = CD = 6, AD = BC = 8

2STEP 2

Subproblem 1 — the base: M halves AD = 8, so the base AM = 4.

AM =

\frac{1}{2}

× AD =

\frac{1}{2}

× 8 = 4

3STEP 3

Subproblem 2 — the height: DC ⊥ AD, so the height is DC = 6 (equal to AB).

h = DC = AB = 6

4STEP 4

Subproblem 3 — the area: ½ × base 4 × height 6 gives the area 12.

Area(△ AMC) =

\frac{1}{2}

× 4 × 6 =

\frac{1}{2}

× 24 = 12 → (A)

Answer

Sanity check by comparing to the whole rectangle. The rectangle has area 6 × 8 = 48. The diagonal AC splits the rectangle into two equal triangles, so △ ACD has area 24. Triangle AMC shares vertex C with △ ACD but uses only half of the base AD (since AM =

\frac{1}{2}

AD), so its area should be half of 24, which is 12. That matches choice (A).

💡Key takeaway

This AMC 8 problem only needs the Grade 6 triangle area formula — pick a smart base where the height is already drawn for you, and the answer falls out in one step.

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