AMC 8 · 2000 · #25

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-rectanglesarea-trianglescoordinate-geometryfraction-multiplication area-differenceidentify-subproblemscoordinate-geometry ↑ Prerequisites: area-rectanglesarea-triangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

The area of rectangle $ABCD$ is $72$ units squared. If point $A$ and the midpoints of $\overline{BC}$ and $\overline{CD}$ are joined to form a triangle, the area of that triangle is

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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Toolkit + CCSS Solution

Understand

Restated: Rectangle $ABCD$ has area $72$. Let $M$ be the midpoint of $\overline{BC}$ and $N$ be the midpoint of $\overline{CD}$. Connect $A$, $M$, and $N$ to form a triangle. Find the area of $\triangle AMN$.

Givens: Rectangle $ABCD$ with area $72$ square units; $M$ is the midpoint of side $\overline{BC}$; $N$ is the midpoint of side $\overline{CD}$; Answer choices: (A) $21$, (B) $27$, (C) $30$, (D) $36$, (E) $40$

Unknowns: The area of $\triangle AMN$

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #1 Draw a Diagram, #9 Solve an Easier Related Problem

Computing the area of $\triangle AMN$ directly is awkward because none of its sides are horizontal or vertical. Tool #16 (Count the Complement) flips the problem: $A$, $M$, $N$ cut three corner right triangles out of rectangle $ABCD$, and the inside triangle is whatever is left. Each corner triangle has legs along the rectangle's sides, so its area is easy. Tool #1 (Draw a Diagram) makes the three corner pieces visible. Tool #9 (Easier Related Problem) lets us pick a convenient rectangle with area $72$ — say $12 \times 6$ — to keep arithmetic with whole numbers; the answer for a midpoint construction depends only on the area, not the specific shape.

Execute — Answer: B

#1 Draw a Diagram 5.G.A.1 Step 1

Draw the rectangle and mark the points.
Place $D$ at the bottom-left, $C$ at the bottom-right, $B$ at the top-right, $A$ at the top-left (matching the figure).
Mark $M$ at the middle of the right side $\overline{BC}$ and $N$ at the middle of the bottom side $\overline{CD}$.
Pick a friendly rectangle with area $72$: let $AB = CD = 12$ and $BC = AD = 6$.
The picture shows three right triangles tucked into the corners $B$, $C$, $D$ — and triangle $AMN$ is the leftover middle piece.

$$AB = 12,\ BC = 6,\ \text{area}(ABCD) = 12 \times 6 = 72$$

💡 A clean rectangle on a grid makes every length a whole number, so each corner triangle's area is just a quick base-times-height-divided-by-two.

#1 Draw a Diagram 5.G.A.1 Step 2

Read off the four key lengths from the diagram.
$M$ is the midpoint of $\overline{BC}$, so $BM = MC = \tfrac{6}{2} = 3$.
$N$ is the midpoint of $\overline{CD}$, so $CN = ND = \tfrac{12}{2} = 6$.

$$BM = MC = 3,\quad CN = ND = 6$$

💡 "Midpoint" simply means "half the side" — once labeled on the picture, the legs of each corner triangle are obvious.

#9 Solve an Easier Related Problem 6.G.A.1 Step 3

Compute the three corner right triangles.
Each has its right angle at a corner of the rectangle, with legs along two sides.\n\n• $\triangle ABM$ has the right angle at $B$, with legs $AB = 12$ and $BM = 3$.
Area $= \tfrac{1}{2}(12)(3) = 18$.\n• $\triangle MCN$ has the right angle at $C$, with legs $MC = 3$ and $CN = 6$.
Area $= \tfrac{1}{2}(3)(6) = 9$.\n• $\triangle AND$ has the right angle at $D$, with legs $ND = 6$ and $AD = 6$.
Area $= \tfrac{1}{2}(6)(6) = 18$.

$$[ABM] = 18,\quad [MCN] = 9,\quad [AND] = 18$$

💡 Right triangles with legs on the rectangle's sides are the easiest area to compute — that's why subtracting them is the right move.

#16 Change Focus / Count the Complement 6.G.A.1 Step 4

Subtract the three corner triangles from the rectangle.
The three corners cover everything outside $\triangle AMN$, so what is left is $\triangle AMN$ itself.

$$[AMN] = 72 - (18 + 9 + 18) = 72 - 45 = 27 \;\Rightarrow\; \textbf{(B)}$$

💡 When the target shape is hard to measure but its surroundings are easy, subtract the surroundings from the whole.

[1] #1 5.G.A.1 Draw the rectangle and mark the points. Place $D$ at the bottom-left, $C$ at the

[2] #1 5.G.A.1 Read off the four key lengths from the diagram. $M$ is the midpoint of $\overlin

[3] #9 6.G.A.1 Compute the three corner right triangles. Each has its right angle at a corner o

[4] #16 6.G.A.1 Subtract the three corner triangles from the rectangle. The three corners cover

Review

Reasonableness: Check the answer is independent of the rectangle's shape. Repeat with $AB = 9$ and $BC = 8$ (still area $72$): $BM = 4$, $CN = 4.5$, $AD = 8$. Then $[ABM] = \tfrac{1}{2}(9)(4) = 18$, $[MCN] = \tfrac{1}{2}(4)(4.5) = 9$, $[AND] = \tfrac{1}{2}(4.5)(8) = 18$. Sum $= 45$, and $[AMN] = 72 - 45 = 27$. The same answer comes out — confirming the area only depends on the rectangle's area. A symbolic check makes this exact: if $AB = l$ and $BC = w$, then $[ABM] = \tfrac{lw}{4}$, $[MCN] = \tfrac{lw}{8}$, $[AND] = \tfrac{lw}{4}$, summing to $\tfrac{5lw}{8}$, and $[AMN] = lw - \tfrac{5lw}{8} = \tfrac{3lw}{8} = \tfrac{3}{8}(72) = 27$. Answer (B).

Alternative: Tool #13 (Convert to Algebra): set up coordinates with $D = (0,0)$, $C = (l, 0)$, $B = (l, w)$, $A = (0, w)$, so $M = (l, w/2)$ and $N = (l/2, 0)$. The Shoelace formula gives $[AMN] = \tfrac{1}{2}|x_A(y_M - y_N) + x_M(y_N - y_A) + x_N(y_A - y_M)| = \tfrac{1}{2}|0 \cdot (w/2) + l(0 - w) + (l/2)(w - w/2)| = \tfrac{1}{2}|{-lw} + lw/4| = \tfrac{3lw}{8} = \tfrac{3}{8}(72) = 27$. Same answer (B), but heavier machinery — the complement-subtraction route is faster for an AMC 8 student.

CCSS standards used (min grade 6)

5.G.A.1 Use a coordinate system to graph points and identify positions in the plane (Placing the rectangle on a grid with whole-number side lengths and locating the midpoints $M$ and $N$ on the sides so every length can be read directly from the figure.)
6.G.A.1 Find the area of right triangles and other polygons by composing or decomposing into shapes (Decomposing the rectangle into $\triangle AMN$ plus three corner right triangles ($\triangle ABM$, $\triangle MCN$, $\triangle AND$), computing each right-triangle area with $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$, and subtracting to recover $[AMN]$.)

⭐ The triangle $AMN$ is tilted, so its area is hard to measure head-on. Flip the question: the three corner triangles $ABM$, $MCN$, $AND$ sit on the rectangle's sides, so their areas are easy ($18 + 9 + 18 = 45$). Subtract from the rectangle's area: $72 - 45 = 27$. Answer (B). The trick — "count what's around it instead of what you want" — works for any area $72$, no matter the rectangle's shape.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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