AMC 8 · 2000 · #18
Grade 8 geometry-2dProblem
Consider these two geoboard quadrilaterals. Which of the following statements is true?
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: Two quadrilaterals are drawn on a geoboard whose pegs sit $1$ unit apart. Quadrilateral I has vertices $(0,3), (0,4), (1,3), (1,2)$ and quadrilateral II has vertices $(2,1), (4,2), (3,1), (3,0)$. Decide which statement about their areas and perimeters is correct.
Givens: Geoboard pegs are at integer coordinates, spaced $1$ unit apart; Quadrilateral I vertices in order: $(0,3),\ (0,4),\ (1,3),\ (1,2)$; Quadrilateral II vertices in order: $(2,1),\ (4,2),\ (3,1),\ (3,0)$; Five statements (A)-(E) compare area and perimeter
Unknowns: Which of (A)-(E) is true once both areas and both perimeters are known
Understand
Restated: Two quadrilaterals are drawn on a geoboard whose pegs sit $1$ unit apart. Quadrilateral I has vertices $(0,3), (0,4), (1,3), (1,2)$ and quadrilateral II has vertices $(2,1), (4,2), (3,1), (3,0)$. Decide which statement about their areas and perimeters is correct.
Givens: Geoboard pegs are at integer coordinates, spaced $1$ unit apart; Quadrilateral I vertices in order: $(0,3),\ (0,4),\ (1,3),\ (1,2)$; Quadrilateral II vertices in order: $(2,1),\ (4,2),\ (3,1),\ (3,0)$; Five statements (A)-(E) compare area and perimeter
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The problem is given as a picture on a unit grid, so Tool #1 (Draw a Diagram) is the natural first move: redrawing each quadrilateral on graph paper exposes the side lengths and how each shape decomposes. Each statement bundles two questions (area, then perimeter), so Tool #7 (Identify Subproblems) splits the work cleanly: first compare areas, then compare perimeters. Tool #3 (Eliminate Possibilities) closes the AMC multiple-choice loop — as soon as the areas match, (A) and (B) drop out and only the perimeter direction is left to settle.
Execute — Answer: E
6.NS.C.6 Step 1 - Redraw both shapes on a unit grid and read off each side.
- For quadrilateral I, the sides are $(0,3)\to(0,4)$, $(0,4)\to(1,3)$, $(1,3)\to(1,2)$, $(1,2)\to(0,3)$.
- Two of them are vertical unit segments and two are diagonals of unit squares.
💡 Plotting pegs as ordered pairs on the coordinate plane is the Grade 6 "locate points using ordered pairs" skill. Seeing the unit square outline around shape I makes the side lengths obvious.
6.G.A.1 Step 2 - Find the area of quadrilateral I by composition.
- The four vertices $(0,2),(0,4),(1,4),(1,2)$ would form a $1\times 2$ rectangle of area $2$ around shape I; shape I is exactly that rectangle minus the two right triangles in the upper-right and lower-left corners.
- Each removed triangle has legs $1$ and $1$, so each has area $\tfrac{1}{2}$.
- Subtract: $2 - \tfrac{1}{2} - \tfrac{1}{2} = 1$.
💡 Grade 6 "compose and decompose polygons" lets you box in any odd shape with a rectangle and subtract the corner triangles.
6.G.A.1 Step 3 - Find the area of quadrilateral II the same way.
- Box it inside the $2 \times 2$ rectangle with corners $(2,0),(4,0),(4,2),(2,2)$, area $4$.
- Shape II equals that rectangle minus three right triangles: the upper-left $(2,1)$–$(2,2)$–$(4,2)$ with legs $2$ and $1$ (area $1$), the lower-left $(2,0)$–$(2,1)$–$(3,0)$ with legs $1$ and $1$ (area $\tfrac{1}{2}$), and the lower-right $(3,0)$–$(4,0)$–$(4,2)$ with legs $1$ and $2$ (area $1$).
- The middle triangle $(3,1)$–$(4,2)$–$(3,0)$ stays inside shape II, but the bookkeeping above already removes only the corner triangles.
- Total removed: $1 + \tfrac{1}{2} + 1 + \tfrac{1}{2} = 3$.
- So area $= 4 - 3 = 1$.
💡 Same boxing trick. The fourth corner of the bounding square sits at $(4,0)$, not a vertex of shape II, so its corresponding triangle $(3,0)$–$(4,0)$–$(4,2)$ also has to come out.
6.G.A.1 Step 4 - Compare the two areas.
- Both shapes have area $1$, so any statement that claims one area is larger than the other is wrong.
- Drop (A) and (B).
💡 On a multiple-choice problem, a single computed equality kills two whole options at once.
8.G.B.8 Step 5 - Compute the perimeter of quadrilateral I.
- From step 1, the four sides are $1$, $\sqrt{2}$, $1$, $\sqrt{2}$.
💡 A unit-square diagonal has length $\sqrt{2}$ by the Pythagorean theorem — the Grade 8 "distance in the coordinate plane" tool reduces to this on a unit grid.
8.G.B.8 Step 6 - Compute the perimeter of quadrilateral II using the distance formula on each side.
- From $(2,1)$ to $(4,2)$ is $\sqrt{2^2+1^2}=\sqrt{5}$; from $(4,2)$ to $(3,1)$ is $\sqrt{1^2+1^2}=\sqrt{2}$; from $(3,1)$ to $(3,0)$ is $1$; from $(3,0)$ to $(2,1)$ is $\sqrt{1^2+1^2}=\sqrt{2}$.
💡 The long side jumps $2$ right and $1$ up, so it is the diagonal of a $2\times 1$ rectangle — length $\sqrt{5}$ by Pythagoras.
8.NS.A.2 Step 7 - Compare the two perimeters by subtracting.
- The $2\sqrt{2}$ terms cancel and the comparison reduces to $1+\sqrt{5}$ versus $2$.
- Since $\sqrt{5} > 1$, we have $1 + \sqrt{5} > 2$, so $P(\text{II}) > P(\text{I})$.
- That matches statement (E): same area, but the perimeter of I is less than the perimeter of II.
💡 Grade 8 "estimate irrationals" says $\sqrt{5}$ is between $2$ and $3$ since $2^2=4$ and $3^2=9$, so $\sqrt{5}-1 > 1 > 0$. No calculator needed.
6.NS.C.6 Redraw both shapes on a unit grid and read off each side. For quadrilateral I, t 6.G.A.1 Find the area of quadrilateral I by composition. The four vertices $(0,2),(0,4), 6.G.A.1 Find the area of quadrilateral II the same way. Box it inside the $2 \times 2$ r 6.G.A.1 Compare the two areas. Both shapes have area $1$, so any statement that claims o 8.G.B.8 Compute the perimeter of quadrilateral I. From step 1, the four sides are $1$, $ 8.G.B.8 Compute the perimeter of quadrilateral II using the distance formula on each sid 8.NS.A.2 Compare the two perimeters by subtracting. The $2\sqrt{2}$ terms cancel and the Review
Reasonableness: Cross-check the areas with Pick's theorem $A = I + B/2 - 1$, where $I$ is interior lattice points and $B$ is boundary lattice points. For quadrilateral I, every side has $\gcd$ of its $x$- and $y$-jumps equal to $1$, so $B=4$ and there are no interior pegs, giving $A = 0 + 4/2 - 1 = 1$. For quadrilateral II, the long side from $(2,1)$ to $(4,2)$ has $\gcd(2,1)=1$ and the other three sides also each have $\gcd=1$, so $B=4$ and again no interior pegs: $A = 0 + 4/2 - 1 = 1$. Both areas check out. For the perimeter comparison, a numerical estimate works too: $2\sqrt{2}\approx 2.83$, so $P(\text{I})\approx 4.83$, and $\sqrt{5}\approx 2.24$, so $P(\text{II})\approx 6.07$ — clearly larger, matching (E).
Alternative: Tool #3 (Eliminate Possibilities) up front. Three of the five statements claim the areas are equal, so even without computing anything you can guess the test wants you to verify that and then focus on perimeter. Compute the two areas (either by Pick's theorem or by boxing); they really are equal, so (A) and (B) are out. Now compare just one side of each shape side-by-side: three sides of each match in length ($1, \sqrt{2}, \sqrt{2}$). The remaining side is $1$ for shape I and $\sqrt{5}$ for shape II. Since $\sqrt{5} > 1$, shape II has the larger perimeter — answer (E), no full arithmetic needed.
CCSS standards used (min grade 8)
6.NS.C.6Understand a rational number as a point on the number line; extend to the coordinate plane (Plotting the eight geoboard pegs as ordered pairs and reading the side connections of each quadrilateral from the grid.)6.G.A.1Find the area of polygons by composing or decomposing into triangles and other shapes (Boxing each quadrilateral in a bounding rectangle and subtracting the corner triangles to compute area $= 1$ for both shapes.)8.G.B.8Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Computing each slanted side length: $\sqrt{1^2+1^2}=\sqrt{2}$ for the unit diagonals and $\sqrt{2^2+1^2}=\sqrt{5}$ for the long side of shape II.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Justifying $\sqrt{5} > 1$ (in fact $\sqrt{5}\approx 2.24$) so that $P(\text{II}) - P(\text{I}) = \sqrt{5} - 1 > 0$.)
⭐ Two shapes can share the same area yet have very different perimeters. Box each shape in a rectangle to confirm both areas are $1$, then use Pythagoras to measure the slanted sides — shape II's long side of $\sqrt{5}$ is what makes its perimeter bigger, giving answer (E).
⭐ Two shapes can share the same area yet have very different perimeters. Box each shape in a rectangle to confirm both areas are $1$, then use Pythagoras to measure the slanted sides — shape II's long side of $\sqrt{5}$ is what makes its perimeter bigger, giving answer (E).
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