AMC 10 · 2003 · #20

Grade 8 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Vertices then Area
coordinate-geometryarea-trianglessimilar-triangles convert-to-algebra ↑ Prerequisites: coordinate-geometry
📏 Medium solution 💡 2 insights 📊 Diagram
Problem
A rectangle ABCD has a bottom side AB=5 and a side BC=3. On the top side DC, mark F so that it sits 1 away from D, and G so that it sits 2 away from C. Draw line AF and line BG; they meet at a point E above the rectangle. Find the area of triangle AEB, whose base is the full bottom side AB.

Pick an answer.

(A)
10
(B)
$\frac{21}{2}$
(C)
12
(D)
$\frac{25}{2}$
(E)
15

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

The hard part of this problem is locating E, a point that lives outside the rectangle where two slanted lines cross. The cleanest way to pin down a crossing point is to drop the whole figure onto a coordinate grid (Tool #1), so every corner becomes an (x,y) pair and each line becomes an equation. Put the bottom-left corner A at the origin; then A,B,C,D,F,G all get exact coordinates. Next, introduce the two lines as algebra (Tools #4 and #13): line AF and line BG each become a simple y = mx+b. Finding E is then one small subproblem (Tool #7) — set the two equations equal and solve. The area is a second small subproblem: triangle AEB sits on base AB along the x-axis, so its height is just the y-coordinate of E, and area =1/2 · base · height.

1STEP 1

Put the rectangle on a grid

Set A=(0,0) with AB on the x-axis: B=(5,0), C=(5,3), D=(0,3), and on the top side F=(1,3), G=(3,3).

A=(0,0), B=(5,0), C=(5,3), D=(0,3), F=(1,3), G=(3,3)
2STEP 2

Write the equation of line AF

Line AF runs from A=(0,0) to F=(1,3), so its slope is 3/1=3; passing through the origin, it is y=3x.

slope_AF=(3-0)/(1-0)=3 → y=3x
3STEP 3

Write the equation of line BG

Line BG runs from B=(5,0) to G=(3,3), so its slope is 3/(-2)=-3/2 and through B the line is y=-3/2(x-5).

slope_BG=(3-0)/(3-5)=-3/2 → y=-3/2(x-5)
4STEP 4

Find E where the lines meet

Both hold at the crossing, so 3x=-3/2(x-5) gives 9/2 x=15/2, x=5/3, and y=3·5/3=5: E=(5/3,5).

3x=-3/2(x-5) → 9/2 x=15/2 → x=5/3, y=3·5/3=5
5STEP 5

Compute the area of triangle AEB

Base AB=5 lies on the x-axis and the height is E's y-coordinate 5, so the area is 1/2·5·5=25/2, choice (D).

[△ AEB]=1/2 · AB · y_E=1/2·5·5=25/2 → (D)
Answer
25/2
The apex E has height 5, comfortably above the rectangle's top edge at height 3 — which fits the picture, since two lines that are still FG=2 apart at height 3 must climb a bit more before meeting. Sanity-check the height against similar triangles: the top gap is FG = DC-DF-GC = 5-1-2=2. Triangles EFG and EAB are similar with ratio FG:AB=2:5, so the height relation (h-3)/h=2/5 gives h=5 — the same height found with coordinates, so the area 25/2 is confirmed. The value also lands between choices (C) 12 and (E) 15, right where an apex just above the box should put it.
💡Key takeaway

Drop the picture onto a coordinate grid: turn each line into an equation, solve them together to find where they cross, and the height of that crossing point is all you need for the triangle's area.

  • Put the rectangle on a grid
  • Write the equation of line AF
  • Write the equation of line BG
  • Find E where the lines meet
  • Compute the area of triangle AEB

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