AMC 10 · 2003 · #14
Grade 8 geometry-2d
Pick an answer.
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.
Put the rectangle on a grid
Placing a corner at the origin gives every point coordinates, with F and G on the top side.
Giving every corner an (x,y) address turns "where do the lines cross?" into arithmetic instead of guesswork.
5.G.A.1Draw A DiagramWrite the equation of line AF
The first line passes through the origin, so it is simply y = 3x.
A line through the origin is just y = (slope) x, so its rule takes one step to write.
8.EE.B.6Introduce A VariableWrite the equation of line BG
The second has slope negative three halves through the far corner, giving its equation.
Two points fix a line; slope plus one point it passes through is all its equation needs.
8.EE.B.6Convert To AlgebraFind E where the lines meet
Solving the pair puts the crossing at height 5, above the rectangle.
The one point that sits on both lines is found by making their two y-rules agree.
8.EE.C.8Identify SubproblemsCompute the area of triangle AEB
Half of base five times that height gives 25/2, choice (D).
When a triangle's base sits on the x-axis, its height is just how high the top vertex reaches.
With the base sitting on the axis, the triangle's height is just how high the top vertex reaches.
▸ Why?
A triangle covers half of the rectangle on the same base and height, so its area is half that product.
▸ Why?
The vertex was found as the one point satisfying both line equations, so its coordinates are trustworthy on both counts.
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