AMC 10 · 2018 · #18
Grade 8 geometry-2dPick an answer.
Tool #7 (Identify Subproblems): the quadrilateral FDBG is awkward to attack head-on, but it is exactly the big triangle ABG with the small corner triangle ADF removed. So the job splits into two easy areas. Tool #1 (Draw a Diagram) pins down where F and G sit. Tool #4 (Introduce a Variable) handles the angle-bisector split of BC with a ratio variable. Tool #16 (Change Focus / Count the Complement) is the finishing move: instead of measuring FDBG directly, subtract the corner triangle from the larger triangle.
Draw and label the figure
The three given values fix the figure.
A clear picture shows the region is a corner cut off a triangle.
7.G.A.2Draw A DiagramReframe the quadrilateral
See it as a difference of two triangles.
A four-sided region is often just a triangle with a corner snipped off.
6.G.A.1Identify SubproblemsFind [ABG] with the angle bisector
The bisector splits the base in the ratio of the sides.
Same height means area splits in the same ratio as the base.
With the same height, the area splits in exactly the same ratio as the base.
▸ Why?
Triangles sharing an apex over one line share a height, so only the bases can differ.
▸ Why?
An area is half the base times the height, so with the height fixed the area rides on the base alone.
Find [ADF] by scaling
The midsegment makes a half-scale copy.
Halving every length quarters the area.
8.G.A.4Introduce A VariableSubtract to get the quadrilateral
Subtracting gives 75.
The region you want is everything left after the corner is gone.
6.G.A.1Change Focus Count The ComplementTo get a weird four-sided region, find a triangle around it and just subtract the corner you do not want.
- Draw and label the figure
- Reframe the quadrilateral
- Find [ABG] with the angle bisector
- Find [ADF] by scaling
- Subtract to get the quadrilateral