AMC 8 · 2006 · #5
Grade 5 geometry-2d
Pick an answer.
AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The picture is already given, but it pays off to add two more lines: the diagonals AC and BD of the smaller square. Those diagonals are horizontal and vertical, and they split the larger square into 8 small right triangles that are all congruent by the symmetry of midpoints. Once we see those 8 equal triangles, the answer comes from counting: 4 of them tile the smaller square and 4 tile the leftover corners, so the smaller square is exactly half of the larger. No algebra, no Pythagorean theorem.
Draw the inner square's two diagonals AC and BD; they cross at the center of the larger square.
Drawing the diagonals of a square is a Grade 4 "draw lines and segments" move, and it turns this picture into something we can count.
4.G.A.1Draw A DiagramThese cuts plus the inner square's sides slice the larger square into 8 congruent right triangles.
Rotating the picture 90° around the center sends each triangle to another one, so they must all have the same area.
4.G.A.3Analyze The UnitsThe two diagonals cut the inner square into 4 triangles; the other 4 fill the corners, so it is half the larger square.
When equal pieces fill a whole, counting how many you have gives you the fraction — Grade 3 area-as-equal-parts reasoning.
3.G.A.2Draw A DiagramHalf of 60 gives the inner square's area, 30, which is choice (D).
Multiplying a whole-number area by the fraction is the Grade 5 "fraction of a quantity" step.
5.NF.B.4Analyze The UnitsWhen midpoints of a square's sides are joined, the inner square is always half the area of the outer one. Adding two diagonals to the picture makes that fact countable instead of computable.