AMC 10 · 2013 · #13
Grade 8 geometry-2dPick an answer.
The heart of the problem is the unknown crossing point on CD, so Tool #4 (Introduce a Variable) is primary: call the point E=(x,y) and turn "equal areas" into an equation for it. Tool #1 (Draw a Diagram) is used first to see that the line from A to a point on CD cuts off triangle AED sitting on the x-axis. Tool #7 (Identify Subproblems) splits the work into two clean pieces — find the whole area, then match half of it. Tool #13 (Convert to Algebra) writes side CD as a line equation so the point's x-coordinate can be solved once its height is known.
Plot the points and the cut
One side lies along an axis.
Seeing which two pieces the line makes tells you exactly which area to control.
6.G.A.3Draw A DiagramFind the whole area
The whole area comes out to 15/2.
The shoelace formula reads the area straight off the corner coordinates.
6.G.A.1Identify SubproblemsSet the triangle's area to half
Half of it fixes the crossing point's height.
Because the base sits on the axis, the point's height alone sets the triangle's area.
Because the base sits on the axis, the point's height alone sets the triangle's area.
▸ Why?
An area is half the base times the height, and the base is already fixed by the two known corners.
▸ Why?
With that base shared, two triangles' areas differ only by their heights, so the height is the single dial.
Locate the point on side CD
The side's line then gives the other coordinate.
The point must obey the line of CD, so its known height pins down its x.
8.EE.B.6Convert To AlgebraAdd the reduced parts
Adding the parts gives 58, choice (B).
Once the fractions are fully reduced, just add the four whole numbers.
6.NS.C.6Introduce A VariableSince AD lies flat on the x-axis, the cut's triangle has area just 2 × its height, so make that equal half the total, then ride the line of CD to find the exact point.
- Plot the points and the cut
- Find the whole area
- Set the triangle's area to half
- Locate the point on side CD
- Add the reduced parts