AMC 10 · 2013 · #1
Grade 6 geometry-2d
Pick an answer.
Tool #1 (Draw a Diagram) is primary because the whole problem turns on reading the figure correctly: the square's corner at B is a right angle, and since E sits on BC, triangle ABE is a right triangle whose two legs are the side AB and the piece BE. Once the diagram shows those perpendicular legs, Tool #4 (Introduce a Variable) finishes the job: call BE the unknown, write the right-triangle area as 1/2 · AB · BE, set it equal to 40, and solve the one-step equation.
Read the figure: a right triangle at B
The corner gives a right angle for free.
The corner of a square is a right angle, so the two square-edges meeting there are the perpendicular legs of the triangle.
6.G.A.1Draw A DiagramWrite the area with BE as the base
The unknown segment serves as the base.
In a right triangle one leg is the base and the other leg is the height, so the area is just half their product.
6.G.A.1Introduce A VariableSolve for BE
Solving gives 8, choice (E).
To undo a multiplication by 5, divide both sides by 5.
To undo a multiplication by five you divide both sides by five.
▸ Why?
Dividing undoes multiplying exactly, so it strips the coefficient off the unknown.
▸ Why?
Doing the same thing to both sides keeps the equation true, so the step is legitimate.
At a square's corner the two edges are perpendicular, so the triangle is right-angled: its area is half of one leg times the other, and setting that equal to 40 gives BE = 8.
- Read the figure: a right triangle at B
- Write the area with BE as the base
- Solve for BE