Competition · AMC preparation · step 4 of 4
AMC 8 · 2000 · #18
Grade 8 geometry-2d
Pick an answer.
AMC 8 2000 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Read the sides off the grid
Redraw shape I: two sides are vertical unit segments, two are unit-square diagonals, so its sides are 1, √(2), 1, √(2).
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.NS.C.6Draw A DiagramFind the area of shape I
Box shape I in a 1×2 rectangle of area 2, then cut the two corner triangles (each ): its area is 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.1Identify SubproblemsFind the area of shape II
Box shape II in a 2×2 rectangle of area 4 and cut the corner triangles (1++1+=3), so its area is 1 too.
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.1Identify SubproblemsCompare the two areas
Both areas equal 1, so any statement claiming one area is larger is false — drop (A) and (B).
On a multiple-choice problem, a single computed equality kills two whole options at once.
6.G.A.1Eliminate PossibilitiesFind the perimeter of shape I
Add shape I's sides 1 + √(2) + 1 + √(2), so its perimeter is 2 + 2√(2).
A unit-square diagonal has length √(2) by the Pythagorean theorem — the Grade 8 "distance in the coordinate plane" tool reduces to this on a unit grid.
8.G.B.8Identify SubproblemsFind the perimeter of shape II
Apply the distance formula to each side of II: √(5), √(2), 1, √(2), so its perimeter is 1 + 2√(2) + √(5).
The long side jumps 2 right and 1 up, so it is the diagonal of a 2 × 1 rectangle — length √(5) by Pythagoras.
8.G.B.8Identify SubproblemsCompare the two perimeters
Subtract: the 2√(2) terms cancel, leaving 1+√(5) vs 2; since √(5)>1, P(II) > P(I) — matching (E).
Grade 8 "estimate irrationals" says √(5) is between 2 and 3 since 2²=4 and 3²=9, so √(5)-1 > 1 > 0. No calculator needed.
Quadrilateral II has a greater perimeter than quadrilateral I, even though the two shapes have equal area.
▸ Why?
Each perimeter is four side lengths added together, and three of shape I's sides match three of shape II's sides in length, so those matched lengths cancel and the comparison comes down to the single leftover side of each shape.
▸ Why?
A quadrilateral's boundary is its four sides joined end to end with no gaps or overlaps, so the perimeter is exactly those four side lengths added up.
▸ Why?
Shape I's sides measure 1, √(2), 1, √(2) and shape II's measure √(5), √(2), 1, √(2), so three of them (1, √(2), √(2)) pair off one for one as equal lengths and add the same amount to each perimeter.
▸ Why?
Shape I's leftover side is a unit segment of length exactly 1, while shape II's leftover side runs 2 across and 1 up, so it is the hypotenuse of a right triangle with legs 2 and 1; squaring and adding the legs gives 2²+1²=5, so that side has length √(5), which is longer than shape I's leftover 1 — making shape II's leftover the larger contribution.
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).
- Read the sides off the grid
- Find the area of shape I
- Find the area of shape II
- Compare the two areas
- Find the perimeter of shape I
- Find the perimeter of shape II
- Compare the two perimeters
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