AMC 8 · 2001 · #11
Grade 6 geometry-2dProblem
Points A, B, C and D have these coordinates: A(3,2), B(3,−2), C(−3,−2) and D(−3,0). The area of quadrilateral ABCD is
Pick an answer.
AMC 8 2001 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: Four points are plotted on the coordinate plane: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$, and $D(-3,0)$. Connect them in order to form quadrilateral $ABCD$ and find its area.
Givens: Vertices: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$, $D(-3,0)$; The sides are $AB$, $BC$, $CD$, and $DA$ taken in order; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$
Unknowns: The area of quadrilateral $ABCD$
Understand
Restated: Four points are plotted on the coordinate plane: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$, and $D(-3,0)$. Connect them in order to form quadrilateral $ABCD$ and find its area.
Givens: Vertices: $A(3,2)$, $B(3,-2)$, $C(-3,-2)$, $D(-3,0)$; The sides are $AB$, $BC$, $CD$, and $DA$ taken in order; Answer choices: (A) $12$, (B) $15$, (C) $18$, (D) $21$, (E) $24$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
Coordinates always invite Tool #1 (Draw a Diagram): plot the four points on grid paper and the quadrilateral's shape becomes obvious. Once we can see it, the slanted side $DA$ is the only awkward feature, so we reach for Tool #7 (Identify Subproblems): cut the quadrilateral along the $x$-axis into a rectangle (below) and a right triangle (above). Both have horizontal and vertical sides, so each area is one easy Grade 6 formula. Add the two pieces. We deliberately skip Tool #13 (algebra/coordinate formulas like the Shoelace formula) because decomposition is faster and uses only elementary area formulas.
Execute — Answer: C
5.G.A.2 Step 1 - Plot the four points on the grid and connect them in order $A \to B \to C \to D \to A$.
- The sides $AB$, $BC$, $CD$ all run along grid lines; only $DA$ is slanted.
- The shape is a trapezoid whose two parallel sides ($AB$ and $CD$) are both vertical.
💡 Plotting ordered pairs in the coordinate plane and reading off the resulting figure is the core Grade 5 coordinate-graphing skill.
6.G.A.1 Step 2 - Use the $x$-axis as a cut line.
- It passes through $D(-3,0)$ and the point $(3,0)$ on side $DA$.
- The cut splits $ABCD$ into two clean pieces: a rectangle below the $x$-axis and a right triangle above it.
💡 Decomposing a polygon into rectangles and right triangles is exactly the Grade 6 "find area by composing/decomposing" technique.
4.MD.A.3 Step 3 - Find the rectangle's area.
- Its width is the horizontal distance from $x=-3$ to $x=3$, which is $6$.
- Its height is the vertical distance from $y=-2$ to $y=0$, which is $2$.
💡 Width $\times$ height for a rectangle is the Grade 4 area formula.
6.G.A.1 Step 4 - Find the right triangle's area.
- Its horizontal leg runs from $(-3,0)$ to $(3,0)$ (length $6$), and its vertical leg runs from $(3,0)$ to $(3,2)$ (length $2$).
💡 Half of base times height for a right triangle is the standard Grade 6 area rule.
6.G.A.1 Step 5 Add the two pieces, since they share only the cut segment and do not overlap.
💡 Area is additive over non-overlapping pieces — the heart of the decomposition standard.
5.G.A.2 Plot the four points on the grid and connect them in order $A \to B \to C \to D 6.G.A.1 Use the $x$-axis as a cut line. It passes through $D(-3,0)$ and the point $(3,0) 4.MD.A.3 Find the rectangle's area. Its width is the horizontal distance from $x=-3$ to $ 6.G.A.1 Find the right triangle's area. Its horizontal leg runs from $(-3,0)$ to $(3,0)$ 6.G.A.1 Add the two pieces, since they share only the cut segment and do not overlap. Review
Reasonableness: Bounding-box check: the smallest rectangle that contains all four points spans $x \in [-3,3]$ and $y \in [-2,2]$, an area of $6 \times 4 = 24$. Our answer $18$ sits below $24$, as it must, and the missing piece is a right triangle with legs $6$ and $2$ (area $6$). Indeed $24 - 6 = 18$. The two views agree, so the answer is consistent.
Alternative: Tool #16 (Change Focus / Complement): instead of adding two pieces, start with the bounding rectangle of area $24$ and subtract the only piece of it that lies outside $ABCD$ — the right triangle with vertices $(-3,0),(-3,2),(3,2)$, which has area $\tfrac{1}{2}(6)(2) = 6$. Then $24 - 6 = 18$, the same answer.
CCSS standards used (min grade 6)
5.G.A.2Represent real world and mathematical problems by graphing points in the first quadrant of the coordinate plane (Plotting the four labeled vertices on the coordinate grid so the shape of quadrilateral $ABCD$ becomes visible.)4.MD.A.3Apply the area and perimeter formulas for rectangles in real world and mathematical problems (Computing the rectangle piece's area as width $\times$ height $= 6 \times 2 = 12$.)6.G.A.1Find the area of right triangles and other polygons by composing into rectangles or decomposing into triangles (Cutting $ABCD$ along the $x$-axis into a rectangle and a right triangle, computing each area, and adding them to get $18$.)
⭐ Plot the points, then slice the shape into a rectangle and a right triangle — both areas are Grade 6 formulas, and their sum is the answer.
⭐ Plot the points, then slice the shape into a rectangle and a right triangle — both areas are Grade 6 formulas, and their sum is the answer.
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