AMC 10 · 2011 · #3
Grade 5 geometry-2dPick an answer.
AMC 10 2011 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question asks for a minimum, so Tool #14 (Extreme Principle) drives the solution: push each measurement to the smallest value its reported range allows. Tool #7 (Identify Subproblems) splits the job into two easy pieces — find the smallest each side can be, then multiply. Tool #8 (Analyze the Units) confirms that inches times inches gives square inches, so the number we compute really is an area.
Find the smallest each side can be
A label can be off by 0.5 inch either way, so the smallest true sides are 1.5 in and 2.5 in.
To make the tile as small as possible, shrink every side to the lowest value its label allows.
5.NBT.B.7Extreme PrincipleSmallest sides give the smallest area
Area is a product of two positive lengths, so it is smallest exactly when both sides are at their smallest.
Multiplying by a smaller number always scales the result down, so the two smallest sides give the smallest area.
Multiplying by a smaller number always scales the result down, so the smallest sides give the smallest area.
▸ Why?
With one side fixed, the area rises and falls in step with the other side.
▸ Why?
Shrinking one side then the other keeps the area no larger at each step, so the bottom is reached.
Set up the minimum area
So the minimum area is 1.5 in × 2.5 in, and inches × inches gives square inches, just as an area should.
A rectangle's area is just its two side lengths multiplied together.
4.MD.A.3Analyze The UnitsMultiply the decimals
Split it up: 1.5×2.5=1.5×2+1.5×0.5=3+0.75=3.75 square inches, choice (A).
Split the multiplication into a whole part and a half part, then add.
5.NBT.B.7Analyze The UnitsTo make a product as small as possible, make each thing you multiply as small as it is allowed to be.
- Find the smallest each side can be
- Smallest sides give the smallest area
- Set up the minimum area
- Multiply the decimals