AMC 8 · 2000 · #18

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Lattice or Non-Aligned Polygon Area

coordinate-geometryarea-trianglesperimeterpythagorean-theorem coordinate-geometryidentify-subproblemscasework ↑ Prerequisites: coordinate-geometrypythagorean-theorem

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

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.

Pick an answer.

(A)

The area of quadrilateral I is more than the area of quadrilateral II.

(B)

The area of quadrilateral I is less than the area of quadrilateral II.

(C)

The quadrilaterals have the same area and the same perimeter.

(D)

The quadrilaterals have the same area, but the perimeter of I is more than the perimeter of II.

(E)

The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II.

AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Redraw shape I: two sides are vertical unit segments, two are unit-square diagonals, so its sides are 1, √(2), 1, √(2).

I sides: 1, √(2), 1, √(2)

2STEP 2

Box shape I in a 1×2 rectangle of area 2, then cut the two corner triangles (each $\frac{1}{2}$ ): its area is 1.

Area(I) = 1 · 2 - 2 ·

\frac{1}{2}

· 1 · 1 = 1

3STEP 3

Box shape II in a 2×2 rectangle of area 4 and cut the corner triangles (1+ $\frac{1}{2}$ +1+ $\frac{1}{2}$ =3), so its area is 1 too.

Area(II) = 2 · 2 - 3 = 1

4STEP 4

Both areas equal 1, so any statement claiming one area is larger is false — drop (A) and (B).

Area(I) = Area(II) = 1 → eliminate (A), (B)

5STEP 5

Add shape I's sides 1 + √(2) + 1 + √(2), so its perimeter is 2 + 2√(2).

P(I) = 1 + √(2) + 1 + √(2) = 2 + 2√(2)

6STEP 6

Apply the distance formula to each side of II: √(5), √(2), 1, √(2), so its perimeter is 1 + 2√(2) + √(5).

P(II) = √(5) + √(2) + 1 + √(2) = 1 + 2√(2) + √(5)

7STEP 7

Subtract: the 2√(2) terms cancel, leaving 1+√(5) vs 2; since √(5)＞1, P(II) ＞ P(I) — matching (E).

P(II) - P(I) = (1+√(5)) - 2 = √(5) - 1 ＞ 0 → (E)

Answer

The quadrilaterals have the same area, but the perimeter of I is less than the perimeter of II.

Cross-check the areas with Pick's theorem A = I +

\frac{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 +

\frac{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 +

\frac{4}{2}

- 1 = 1. Both areas check out. For the perimeter comparison, a numerical estimate works too: 2√(2)≈ 2.83, so P(I)≈ 4.83, and √(5)≈ 2.24, so P(II)≈ 6.07 — clearly larger, matching (E).

💡Key takeaway

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 √(5) is what makes its perimeter bigger, giving answer (E).

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