AMC 8 · 2002 · #16
Grade 8 geometry-2d
Pick an answer.
AMC 8 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The setup — three shapes built on the sides of a right triangle, one per side — is the picture that appears in every proof of the Pythagorean theorem. Tool #10 (Use a Related Problem) tells us to lean on 3² + 4² = 5² instead of grinding through five answer choices. Each outer triangle on a side of length s has area s², so the three outer areas are exactly half the three squared sides — multiply Pythagoras by and the equation X+Y=Z pops out. Tool #7 (Break Into Subproblems) handles the bookkeeping: compute W, X, Y, Z one at a time, then test the choices.
Inner 3-4-5 triangle: legs 3 and 4 form the right angle, so its area is W = 6.
Grade 6 "area of a right triangle = base × height" applied to the legs.
6.G.A.1Identify SubproblemsEach outer 45-45-90 triangle has both legs equal to its shared side s, so area s²: X = 4.5, Y = 8, Z = 12.5.
A right isosceles triangle with legs s has area s · s = s² — half of the square on that side.
6.G.A.1Identify SubproblemsThe inner triangle is 3-4-5, so 3² + 4² = 5²; halve both sides and the outer areas satisfy X + Y = Z.
Grade 8 "apply the Pythagorean theorem" — three similar shapes on the sides of a right triangle always satisfy outer_a + outer_b = outer_c, because all three are the same fraction of the squares on the sides.
8.G.B.7Create A Physical RepresentationSubstituting W=6, X=4.5, Y=8, Z=12.5, only choice (E) holds: X + Y = 12.5 = Z.
Grade 6 "check which value makes an equation true" — only (E) survives the numerical test, matching the Pythagorean argument.
6.EE.B.5Identify SubproblemsThree triangles built on a 3-4-5 right triangle is just the Pythagorean picture in disguise — each area is half of a square on a side, so 3²+4²=5² becomes X+Y=Z.