AMC 8 · 2003 · #6
Grade 8 geometry-2d
Pick an answer.
AMC 8 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The figure pairs each square with one side of the triangle, so Tool #7 (Break Into Subproblems) splits the problem into three clean parts: (1) turn each square's area into a side length, (2) check what kind of triangle has sides 5, 12, 13, (3) compute its area. Tool #10 (Use a Related Problem) is the recognition that 5-12-13 is a famous Pythagorean triple, so the converse of the Pythagorean theorem from a related problem makes step (3) easy — the triangle is right-angled, and its legs are the base and height.
Take the square root of each area to get its square's side: 5, 12, and 13.
Grade 8 "use square root to solve x² = p" — each side is the square root of its square's area.
8.EE.A.2Identify SubproblemsEach triangle side is a shared square side, so the triangle has sides 5, 12, 13.
Grade 7 "draw geometric shapes with given conditions" — sides of the squares are exactly the sides of the triangle.
7.G.A.2Identify SubproblemsSince 5² + 12² = 169 = 13², the converse of the Pythagorean theorem makes it a right triangle with legs 5 and 12.
Grade 8 "explain a proof of the converse of the Pythagorean theorem" — 5-12-13 is the classic right-triangle pattern.
8.G.B.6Create A Physical RepresentationThe legs are the base and height, so the area is half their product: · 5 · 12 = 30.
Grade 6 "find area of right triangles" — once base and height are the legs, the formula gives the answer in one step.
6.G.A.1Identify SubproblemsEach square's area gives you a side of the triangle. Spot the 5-12-13 right triangle and the area is just · 5 · 12 = 30.