AMC 10 · 2003 · #20
Grade 8 geometry-2dIn rectangle ABCD,AB=5 and BC=3. Points F and G are on CD so that DF=1 and GC=2. Lines AF and BG intersect at E. Find the area of △AEB.
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $AB$.
Givens: $ABCD$ is a rectangle with $AB=5$ and $BC=3$.; $F$ and $G$ lie on the top side $\overline{DC}$ with $DF=1$ and $GC=2$.; $E$ is the intersection of line $AF$ and line $BG$.; Answer choices: (A) $10$, (B) $\tfrac{21}{2}$, (C) $12$, (D) $\tfrac{25}{2}$, (E) $15$.
Unknowns: The area of triangle $AEB$.
Understand
Restated: 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 $AB$.
Givens: $ABCD$ is a rectangle with $AB=5$ and $BC=3$.; $F$ and $G$ lie on the top side $\overline{DC}$ with $DF=1$ and $GC=2$.; $E$ is the intersection of line $AF$ and line $BG$.; Answer choices: (A) $10$, (B) $\tfrac{21}{2}$, (C) $12$, (D) $\tfrac{25}{2}$, (E) $15$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #13 Convert to Algebra, #7 Identify Subproblems
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 $=\tfrac12 \cdot \text{base} \cdot \text{height}$.
Execute — Answer: D
5.G.A.1 Step 1 Put the rectangle on a grid
- Place the bottom-left corner $A$ at the origin and let $AB$ run along the $x$-axis.
- Since $AB=5$ and $BC=3$, the four corners are $A=(0,0)$, $B=(5,0)$, $C=(5,3)$, $D=(0,3)$.
- The top side $DC$ is the line $y=3$, running from $D=(0,3)$ to $C=(5,3)$.
- Now place $F$ and $G$ on that top side.
- $F$ is $1$ to the right of $D$, so $F=(1,3)$.
- $G$ is $2$ to the left of $C$, so $G=(5-2,\,3)=(3,3)$.
💡 Giving every corner an $(x,y)$ address turns "where do the lines cross?" into arithmetic instead of guesswork.
8.EE.B.6 Step 2 Write the equation of line AF
- Line $AF$ goes from $A=(0,0)$ to $F=(1,3)$.
- Its slope is the rise over the run: $\frac{3-0}{1-0}=3$.
- Because it passes through the origin, its equation is simply $y=3x$.
💡 A line through the origin is just $y = (\text{slope})\,x$, so its rule takes one step to write.
8.EE.B.6 Step 3 Write the equation of line BG
- Line $BG$ goes from $B=(5,0)$ to $G=(3,3)$.
- Its slope is $\frac{3-0}{3-5}=\frac{3}{-2}=-\frac32$.
- Using point $B=(5,0)$, the line is $y-0=-\frac32(x-5)$, that is $y=-\frac32 x+\frac{15}{2}$.
💡 Two points fix a line; slope plus one point it passes through is all its equation needs.
8.EE.C.8 Step 4 Find E where the lines meet
- At the crossing point $E$ both equations hold, so set them equal: $3x=-\frac32(x-5)$.
- Multiply out the right side: $3x=-\frac32 x+\frac{15}{2}$.
- Add $\frac32 x$ to both sides: $\frac92 x=\frac{15}{2}$.
- So $x=\frac{15}{2}\cdot\frac{2}{9}=\frac{15}{9}=\frac53$.
- Put this back into $y=3x$: $y=3\cdot\frac53=5$.
- So $E=\left(\frac53,5\right)$.
💡 The one point that sits on both lines is found by making their two $y$-rules agree.
6.G.A.1 Step 5 Compute the area of triangle AEB
- Triangle $AEB$ has base $AB$ lying on the $x$-axis from $A=(0,0)$ to $B=(5,0)$, so the base length is $5$.
- The height is the perpendicular distance from $E$ up to that base, which is exactly the $y$-coordinate of $E$, namely $5$.
- Therefore the area is $\frac12\cdot\text{base}\cdot\text{height}=\frac12\cdot5\cdot5=\frac{25}{2}$.
- That is choice (D).
💡 When a triangle's base sits on the $x$-axis, its height is just how high the top vertex reaches.
5.G.A.1 Place the bottom-left corner $A$ at the origin and let $AB$ run along the $x$-ax 8.EE.B.6 Line $AF$ goes from $A=(0,0)$ to $F=(1,3)$. Its slope is the rise over the run: 8.EE.B.6 Line $BG$ goes from $B=(5,0)$ to $G=(3,3)$. Its slope is $\frac{3-0}{3-5}=\frac{ 8.EE.C.8 At the crossing point $E$ both equations hold, so set them equal: $3x=-\frac32(x 6.G.A.1 Triangle $AEB$ has base $AB$ lying on the $x$-axis from $A=(0,0)$ to $B=(5,0)$, Review
Reasonableness: 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 $\frac{h-3}{h}=\frac{2}{5}$ gives $h=5$ — the same height found with coordinates, so the area $\frac{25}{2}$ is confirmed. The value also lands between choices (C) $12$ and (E) $15$, right where an apex just above the box should put it.
Alternative: Skip coordinates and use similar triangles directly. Because $DC\parallel AB$, triangle $EFG$ (top, small) is similar to triangle $EAB$ (bottom, large). The parallel sides are $FG=5-1-2=2$ and $AB=5$, so the similarity ratio is $\frac{2}{5}$. Let $h$ be $E$'s height above $AB$; the height of the small triangle above $FG$ is $h-3$. Matching the ratio, $\frac{h-3}{h}=\frac{2}{5}$, which solves to $h=5$. Then area $=\frac12\cdot5\cdot5=\frac{25}{2}$, the same answer without ever writing a line equation.
CCSS standards used (min grade 8)
5.G.A.1Use a pair of perpendicular number lines forming a coordinate system (Placing the rectangle and the points $F,G$ at exact coordinates on a grid.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Computing the slopes of lines $AF$ and $BG$ and writing each line as an equation.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the two line equations together to locate the intersection point $E$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the area of triangle $AEB$ from its base and height.)
⭐ 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.
⭐ 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.
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