AMC 8 · 2000 · #25
Grade 6 geometry-2dProblem
The area of rectangle ABCD is 72 units squared. If point A and the midpoints of BC and CD are joined to form a triangle, the area of that triangle is
Pick an answer.
AMC 8 2000 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: 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
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) $40$
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
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.
💡 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.
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 $CN = ND = \tfrac{12}{2} = 6$.
💡 "Midpoint" simply means "half the side" — once labeled on the picture, the legs of each corner triangle are obvious.
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 $= \tfrac{1}{2}(6)(6) = 18$.
💡 Right triangles with legs on the rectangle's sides are the easiest area to compute — that's why subtracting them is the right move.
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.
💡 When the target shape is hard to measure but its surroundings are easy, subtract the surroundings from the whole.
5.G.A.1 Draw the rectangle and mark the points. Place $D$ at the bottom-left, $C$ at the 5.G.A.1 Read off the four key lengths from the diagram. $M$ is the midpoint of $\overlin 6.G.A.1 Compute the three corner right triangles. Each has its right angle at a corner o 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.1Use 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.1Find 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.
⭐ 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.
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