AMC 8 · 2004 · #14

Grade 6 geometry-2d

Archetype Coordinate Plane Figure Computation

coordinate-geometryarea-trianglesspatial-visualization coordinate-geometryidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-triangles

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

What is the area enclosed by the geoboard quadrilateral below?

Pick an answer.

(A)

(B)

18rac12

(C)

22rac12

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: On a $10 \times 10$ geoboard, a quadrilateral is drawn by connecting the pegs $(4,0)$, $(0,5)$, $(3,4)$, $(10,10)$ in order. Find the area enclosed by this quadrilateral.

Givens: Vertices in order: $V_1=(4,0)$, $V_2=(0,5)$, $V_3=(3,4)$, $V_4=(10,10)$; Each peg sits at integer coordinates on a unit grid; Answer choices: (A) $15$, (B) $18\tfrac12$, (C) $22\tfrac12$, (D) $27$, (E) $41$

Unknowns: The area enclosed by the quadrilateral $V_1 V_2 V_3 V_4$

Understand

Restated: On a $10 \times 10$ geoboard, a quadrilateral is drawn by connecting the pegs $(4,0)$, $(0,5)$, $(3,4)$, $(10,10)$ in order. Find the area enclosed by this quadrilateral.

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Break Into Subproblems, #13 Use Algebra

The shape is given by coordinates, so Tool #1 (Draw a Diagram) is the first move: plotting the four pegs reveals the quadrilateral is concave at $V_3$. A concave polygon is messy to handle as one piece, so Tool #7 (Break Into Subproblems) splits it along diagonal $V_1V_3$ into two ordinary triangles. Each triangle's vertices are integer coordinates, so Tool #13 (Use Algebra) finishes the job via the coordinate area formula. Adding the two triangle areas gives the total.

Execute — Answer: C

#1 Draw a Diagram 6.NS.C.6 Step 1

Plot the vertices on graph paper.
The pegs are $V_1=(4,0)$ on the bottom, $V_2=(0,5)$ on the left, $V_3=(3,4)$ near the middle, and $V_4=(10,10)$ at the top-right corner.
Connecting them in order $V_1\to V_2\to V_3\to V_4\to V_1$ shows a four-sided shape with a dent at $V_3$ — it's concave, not convex.

$$V_1=(4,0),\ V_2=(0,5),\ V_3=(3,4),\ V_4=(10,10)$$

💡 Plotting points in all four quadrants of a coordinate plane is the Grade 6 "locate points using ordered pairs" skill. Seeing the shape is the whole reason for drawing it.

#7 Break Into Subproblems 6.G.A.1 Step 2

Cut the quadrilateral along diagonal $V_1V_3$.
Because $V_3$ is the dented (concave) vertex, the diagonal from $V_1$ to $V_3$ stays inside the shape and splits it into two triangles: triangle $A=V_1V_2V_3$ on the left and triangle $B=V_1V_3V_4$ on the right.
The full area is the sum of the two triangle areas.

$$\text{Area}(V_1V_2V_3V_4) = \text{Area}(A) + \text{Area}(B)$$

💡 Grade 6 "composing and decomposing polygons" says any polygon can be broken into triangles. A diagonal through the concave vertex is the natural cut.

#13 Use Algebra 6.G.A.3 Step 3

Compute the area of triangle $A$ with vertices $(4,0)$, $(0,5)$, $(3,4)$ using the coordinate area formula.
Label $(x_1,y_1)=(4,0)$, $(x_2,y_2)=(0,5)$, $(x_3,y_3)=(3,4)$ and substitute.

$$\text{Area}(A) = \tfrac12\,|\,4(5-4) + 0(4-0) + 3(0-5)\,| = \tfrac12\,|4 + 0 - 15| = \tfrac{11}{2}$$

💡 The coordinate area formula is the Grade 6 "polygons in the coordinate plane" tool applied to a triangle. The arithmetic is just multiply, add, take absolute value, halve.

#13 Use Algebra 6.G.A.3 Step 4

Compute the area of triangle $B$ with vertices $(4,0)$, $(3,4)$, $(10,10)$ the same way.
Label $(x_1,y_1)=(4,0)$, $(x_2,y_2)=(3,4)$, $(x_3,y_3)=(10,10)$ and substitute.

$$\text{Area}(B) = \tfrac12\,|\,4(4-10) + 3(10-0) + 10(0-4)\,| = \tfrac12\,|-24 + 30 - 40| = \tfrac{34}{2} = 17$$

💡 Same formula, new vertices. The absolute value soaks up the sign, so it doesn't matter which orientation you traverse the triangle in.

#7 Break Into Subproblems 6.G.A.1 Step 5

Add the two triangle areas to get the total enclosed area.

$$\text{Total} = \tfrac{11}{2} + 17 = \tfrac{11}{2} + \tfrac{34}{2} = \tfrac{45}{2} = 22\tfrac12 \;\Rightarrow\; \textbf{(C)}$$

💡 Decomposition only pays off if you remember to add the pieces back together. The sum matches answer choice (C).

[1] #1 6.NS.C.6 Plot the vertices on graph paper. The pegs are $V_1=(4,0)$ on the bottom, $V_2=(

[2] #7 6.G.A.1 Cut the quadrilateral along diagonal $V_1V_3$. Because $V_3$ is the dented (conc

[3] #13 6.G.A.3 Compute the area of triangle $A$ with vertices $(4,0)$, $(0,5)$, $(3,4)$ using t

[4] #13 6.G.A.3 Compute the area of triangle $B$ with vertices $(4,0)$, $(3,4)$, $(10,10)$ the s

[5] #7 6.G.A.1 Add the two triangle areas to get the total enclosed area.

Review

Reasonableness: Cross-check with Pick's theorem for the whole quadrilateral. The boundary lattice points come from the four edges: $V_1V_2$ has $\gcd(4,5)=1$, $V_2V_3$ has $\gcd(3,1)=1$, $V_3V_4$ has $\gcd(7,6)=1$, and $V_4V_1$ has $\gcd(6,10)=2$, contributing one interior lattice point at $(7,5)$. So $B=4+1=5$ boundary points. Counting interior pegs inside the concave region (using the diagram) gives $I=21$. Pick's theorem: $\text{Area} = I + B/2 - 1 = 21 + 2.5 - 1 = 22.5$. Matches (C). Also, the answer must be a multiple of $\tfrac12$ for any geoboard polygon, ruling out anything that isn't a half-integer.

Alternative: Tool #1 (Draw a Diagram) again, this time enclosing triangle $B$ in a bounding rectangle. Triangle $B$ has vertices $(4,0)$, $(3,4)$, $(10,10)$ fitting inside the rectangle $[3,10]\times[0,10]$ of area $70$. Subtract the three right-triangle corners: bottom-right $(4,0)$–$(10,0)$–$(10,10)$ of area $\tfrac12\cdot 6\cdot 10=30$; bottom-left $(3,0)$–$(4,0)$–$(3,4)$ of area $\tfrac12\cdot 1\cdot 4=2$; top-left $(3,4)$–$(3,10)$–$(10,10)$ of area $\tfrac12\cdot 6\cdot 7=21$. So $\text{Area}(B)=70-30-2-21=17$, matching the formula. Apply the same idea to triangle $A$ to confirm $\tfrac{11}{2}$, then add.

CCSS standards used (min grade 6)

6.NS.C.6 Understand a rational number as a point on the number line; extend to coordinate plane (Plotting the four geoboard pegs as ordered pairs on the coordinate plane in Step 1.)
6.G.A.1 Find the area of polygons by composing or decomposing into triangles and other shapes (Splitting the concave quadrilateral along diagonal $V_1V_3$ into two triangles whose areas add to the total.)
6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices (Using vertex coordinates of each triangle to compute its area via the coordinate area formula $\tfrac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$.)

⭐ When a coordinate shape looks weird or dented, draw it first and split it along a diagonal. Two clean triangles are always easier than one concave quadrilateral, and the area formula handles each piece on its own.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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