AMC 10 · 2009 · #14
Grade 6 geometry-2d
Pick an answer.
The shaded region has an awkward staircase edge, but it becomes easy once broken into pieces. Tool #7 (Identify Subproblems) is primary: fill in the one missing unit square at the bottom-right so the shaded part plus that square is a single clean triangle, then the shaded area is just (triangle) minus (one square). Tool #1 (Draw a Diagram) reads the exact corner coordinates off the figure so the triangle's base and height are known. Tool #4 (Introduce a Variable) keeps c as the unknown, writes the triangle's area in terms of c, and turns the equal-area condition into one linear equation to solve.
Find each region's target area
Each half must have the same target area.
Cutting an area in half means each side is exactly half of the whole.
3.MD.C.6Identify SubproblemsComplete the shaded part to a triangle
Completing the piece to a triangle makes it measurable.
Adding the one missing square straightens the jagged edge into a plain triangle, and area is additive so we just subtract it back.
Adding the one missing square straightens the jagged edge into a plain triangle, which is then corrected back.
▸ Why?
Area is additive, so a piece added on can be taken straight back off without any loss.
▸ Why?
A right triangle is half the rectangle built on its two perpendicular sides, so its area is one product halved.
Write the triangle's area with c
Its area is a simple expression in the unknown.
A right triangle is just half of the rectangle built on its two perpendicular sides.
6.G.A.1Introduce A VariableSet up the equation and solve
Solving gives 2/3, choice (C).
Once the equal-area rule fixes the triangle's area, one equation pins down where the line must start.
6.EE.B.7Introduce A VariableFill in the missing square to turn a jagged shape into a clean triangle, find its area, then subtract the square back.
- Find each region's target area
- Complete the shaded part to a triangle
- Write the triangle's area with c
- Set up the equation and solve