AMC 10 · 2013 · #3
Grade 6 geometry-2d
Pick an answer.
AMC 10 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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
AB and BC meet at corner B at a right angle and E sits on BC, so triangle ABE is right-angled at B with legs AB = 10 and BE.
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
Take BE as the base and AB = 10 as the height: the area is half of BE · 10, that is 5 · BE, and it equals 40.
In a right triangle one leg is the base and the other leg is the height, so the area is just half their product.
In a right triangle one leg is the base and the other is the height, so the area is half their product.
▸ Why?
The two legs meet at a right angle, which is exactly what makes one a height for the other.
▸ Why?
A triangle's area is half its base times its matching height, whichever pair you choose.
Solve for BE
Divide 5 · BE = 40 by 5 to get BE = 8, which lies between 0 and 10, so E really sits on BC. The answer is (E).
To undo a multiplication by 5, divide both sides by 5.
6.EE.B.7Introduce A VariableAt 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