AMC 10 · 2012 · #19
Grade 8 geometry-2dIn rectangle ABCD, AB=6, AD=30, and G is the midpoint of AD. Segment AB is extended 2 units beyond B to point E, and F is the intersection of ED and BC. What is the area of quadrilateral BFDG?
Pick an answer.
AMC 10 2012 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 $AB=6$ and $AD=30$. Point $G$ is the midpoint of side $AD$. Side $AB$ is extended past $B$ by 2 units to a point $E$, and $F$ is where segment $ED$ crosses side $BC$. Find the area of the four-sided region $BFDG$.
Givens: $ABCD$ is a rectangle with $AB=6$ and $AD=30$.; $G$ is the midpoint of $\overline{AD}$.; $E$ is on ray $AB$, 2 units beyond $B$, so $BE=2$.; $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$.
Unknowns: The area of quadrilateral $BFDG$.
Understand
Restated: A rectangle $ABCD$ has $AB=6$ and $AD=30$. Point $G$ is the midpoint of side $AD$. Side $AB$ is extended past $B$ by 2 units to a point $E$, and $F$ is where segment $ED$ crosses side $BC$. Find the area of the four-sided region $BFDG$.
Givens: $ABCD$ is a rectangle with $AB=6$ and $AD=30$.; $G$ is the midpoint of $\overline{AD}$.; $E$ is on ray $AB$, 2 units beyond $B$, so $BE=2$.; $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #4 Introduce a Variable
The figure mixes a rectangle, an extended side, and a crossing line, so the cleanest move is to pin it onto a coordinate grid where every point gets a clear address. Then the work splits into two subproblems: first locate $F$ on side $BC$, then measure the area of $BFDG$. Once the points are placed, $BFDG$ turns out to be a trapezoid, so its area is a single formula.
Execute — Answer: C
6.G.A.3 Step 1 Put the rectangle on a grid
- Place $A$ at the origin with $B$ to its right and $D$ above it.
- Then $A=(0,0)$, $B=(6,0)$, $C=(6,30)$, $D=(0,30)$.
- The midpoint $G$ of $\overline{AD}$ is $(0,15)$.
- Extending $AB$ two units past $B$ gives $E=(8,0)$.
- Side $BC$ is the vertical line $x=6$.
💡 Giving every corner an $(x,y)$ address turns a picture into numbers you can compute with.
8.G.A.5 Step 2 Find F with similar triangles
- Look at triangles $EBF$ and $EAD$.
- Both have the angle at $E$, and $BF$ (part of the line $x=6$) is parallel to $AD$ (the line $x=0$), so the triangles are similar.
- The ratio of their sizes is $EB:EA = 2:8 = 1:4$.
- Since $AD=30$, the matching side $BF$ is one quarter of it: $BF=\tfrac{30}{4}=\tfrac{15}{2}$.
- So $F=(6,\tfrac{15}{2})$.
💡 Two triangles that share an angle and have parallel bases are scaled copies, so their sides shrink by the same factor.
6.G.A.1 Step 3 See BFDG as a trapezoid
- The corners are $B=(6,0)$, $F=(6,\tfrac{15}{2})$, $D=(0,30)$, $G=(0,15)$.
- Side $BF$ lies on the vertical line $x=6$ and side $GD$ lies on the vertical line $x=0$, so $BF$ and $GD$ are parallel.
- That makes $BFDG$ a trapezoid.
- Its two parallel sides are $BF=\tfrac{15}{2}$ and $GD=30-15=15$, and the distance between the two vertical lines is $6$.
💡 When two sides of a four-sided figure are parallel, the whole shape is a trapezoid and one formula gives its area.
6.G.A.1 Step 4 Apply the trapezoid area formula
- The area of a trapezoid is the average of the two parallel sides times the height.
- Here that is $\tfrac12(BF+GD)\cdot 6 = \tfrac12\left(\tfrac{15}{2}+15\right)\cdot 6 = \tfrac12\cdot\tfrac{45}{2}\cdot 6 = \tfrac{135}{2}$.
- So the area of $BFDG$ is $\tfrac{135}{2}$, which is answer (C).
💡 Averaging the two parallel sides turns the slanted trapezoid into an equal-area rectangle you can measure.
6.G.A.3 Place $A$ at the origin with $B$ to its right and $D$ above it. Then $A=(0,0)$, 8.G.A.5 Look at triangles $EBF$ and $EAD$. Both have the angle at $E$, and $BF$ (part of 6.G.A.1 The corners are $B=(6,0)$, $F=(6,\tfrac{15}{2})$, $D=(0,30)$, $G=(0,15)$. Side $ 6.G.A.1 The area of a trapezoid is the average of the two parallel sides times the heigh Review
Reasonableness: The whole rectangle has area $6\times 30=180$. Our trapezoid area $\tfrac{135}{2}=67.5$ is a bit over a third of that, which looks right for a region spanning the full height on one side and part of it on the other. A quick Shoelace check on $B(6,0),F(6,7.5),D(0,30),G(0,15)$ also gives $\tfrac12|135|=\tfrac{135}{2}$, confirming the answer.
Alternative: Instead of the trapezoid formula, drop the coordinates straight into the Shoelace formula for the four vertices $B$, $F$, $D$, $G$; it returns $\tfrac{135}{2}$ directly. Alternatively, find $F$ by writing the line $ED$ as $y=-\tfrac{15}{4}(x-8)$ and setting $x=6$ to get $y=\tfrac{15}{2}$.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the rectangle, the midpoint $G$, and the extended point $E$ on a grid so every corner has coordinates.)8.G.A.5Use informal arguments to establish the angle-angle criterion for similarity of triangles (Recognizing triangles $EBF$ and $EAD$ as similar to find $BF=\tfrac{15}{2}$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Identifying $BFDG$ as a trapezoid and computing its area with the trapezoid formula.)
⭐ Drop the shape onto a grid, use similar triangles to find the missing corner, and the region becomes a trapezoid you can measure with one formula.
⭐ Drop the shape onto a grid, use similar triangles to find the missing corner, and the region becomes a trapezoid you can measure with one formula.
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