AMC 8 · 2008 · #23

Grade 6 geometry-2d

Archetype Compound Area by Decomposition / Difference

area-trianglesarea-rectanglescoordinate-geometryfraction-arithmetic area-differenceidentify-subproblemscoordinate-geometry ↑ Prerequisites: area-trianglesarea-rectangles

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In square $ABCE$ , $AF=2FE$ and $CD=2DE$ . What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$ ?

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{2}{9}$

(C)

$\frac{5}{18}$

(D)

$\frac{1}{3}$

(E)

$\frac{7}{20}$

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Toolkit + CCSS Solution

Understand

Restated: Square $ABCE$ has point $F$ on side $AE$ with $AF = 2 \cdot FE$, and point $D$ on side $CE$ with $CD = 2 \cdot DE$. Find the ratio of the area of $\triangle BFD$ to the area of square $ABCE$.

Givens: $ABCE$ is a square (vertices in order: $A$ top-left, $B$ top-right, $C$ bottom-right, $E$ bottom-left); $F$ lies on side $AE$ with $AF = 2 \cdot FE$; $D$ lies on side $CE$ with $CD = 2 \cdot DE$; Answer choices: (A) $\tfrac{1}{6}$, (B) $\tfrac{2}{9}$, (C) $\tfrac{5}{18}$, (D) $\tfrac{1}{3}$, (E) $\tfrac{7}{20}$

Unknowns: The ratio $\dfrac{[\triangle BFD]}{[ABCE]}$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #16 Change Representation, #7 Break into Subproblems

Tool #1 (Draw a Diagram) is the natural entry: placing the square on a coordinate grid pins down every point in the figure. Tool #16 (Change Representation) lets us pick the side length, and choosing $s=3$ turns the $2{:}1$ ratios into whole-number lengths so no fractions appear until the very last step. Tool #7 (Break into Subproblems) handles $\triangle BFD$ indirectly: instead of computing its area directly, we cut the square into $\triangle BFD$ plus three right triangles in the corners, find those three easy areas, and subtract.

Execute — Answer: C

#1 Draw a Diagram 6.G.A.3 Step 1

Place the square on a coordinate grid with side length $s=3$.
Put $E$ at the origin so the square sits in the first quadrant: $E(0,0)$, $C(3,0)$, $B(3,3)$, $A(0,3)$.
The area of square $ABCE$ is $s^2 = 9$.

$$E=(0,0),\; C=(3,0),\; B=(3,3),\; A=(0,3),\; [ABCE]=9$$

💡 Grade 6 "polygons in the coordinate plane": coordinates make every side length a count of unit squares.

#16 Change Representation 6.RP.A.3 Step 2

Use the $2{:}1$ ratios to locate $F$ and $D$.
Side $AE$ has length $3$, and $AF = 2 \cdot FE$ means $AE$ is split into thirds with $FE = 1$ and $AF = 2$.
So $F$ is $1$ unit up from $E$ on the $y$-axis.
Likewise on side $CE$, $CD = 2 \cdot DE$ gives $DE = 1$ and $CD = 2$, so $D$ is $1$ unit right of $E$ on the $x$-axis.

$FE = 1,\; AF = 2 \;\Rightarrow\; F=(0,1)$; $\;\; DE = 1,\; CD = 2 \;\Rightarrow\; D=(1,0)$

💡 If a length splits in a $2{:}1$ ratio, each $1$ part is $\tfrac{1}{3}$ of the whole. Scaling the square to side $3$ makes those parts plain integers.

#7 Break into Subproblems 6.G.A.1 Step 3

Break the square into $\triangle BFD$ plus three right triangles in the corners at $A$, $C$, $E$.
The square's interior is exactly the union of these four triangles with no overlap (the segments $BF$, $BD$, $FD$ are the cut lines).
So $[\triangle BFD] = [ABCE] - [\triangle ABF] - [\triangle BCD] - [\triangle FED]$.

$$[ABCE] = [\triangle ABF] + [\triangle BCD] + [\triangle FED] + [\triangle BFD]$$

💡 Grade 6 "compose and decompose polygons": the central triangle is whatever the square has left after the three corner pieces are removed.

#7 Break into Subproblems 6.G.A.1 Step 4

Compute the three corner right-triangle areas.
Each has its right angle at a square vertex, with legs along the square's sides.
$\triangle ABF$ has legs $AB = 3$ (top side) and $AF = 2$ (top part of the left side).
$\triangle BCD$ has legs $BC = 3$ (right side) and $CD = 2$ (right part of the bottom).
$\triangle FED$ has legs $FE = 1$ and $ED = 1$ at corner $E$.

$$[\triangle ABF] = \tfrac{1}{2}(3)(2) = 3,\; [\triangle BCD] = \tfrac{1}{2}(3)(2) = 3,\; [\triangle FED] = \tfrac{1}{2}(1)(1) = \tfrac{1}{2}$$

💡 Right triangle area $= \tfrac{1}{2}\cdot\text{leg}\cdot\text{leg}$ — the Grade 6 base-times-height rule for a half-rectangle.

#7 Break into Subproblems 6.RP.A.1 Step 5

Subtract to get the area of $\triangle BFD$, then divide by the square's area to form the ratio.
Simplify by writing $9 = \tfrac{18}{2}$ so the fractions share a denominator.

$[\triangle BFD] = 9 - 3 - 3 - \tfrac{1}{2} = \tfrac{5}{2}$; $\;\; \dfrac{[\triangle BFD]}{[ABCE]} = \dfrac{5/2}{9} = \dfrac{5}{18} \;\Rightarrow\; \textbf{(C)}$

💡 Dividing a fractional area by a whole area is a Grade 6 ratio — multiply top and bottom by $2$ to clear the inner half.

[1] #1 6.G.A.3 Place the square on a coordinate grid with side length $s=3$. Put $E$ at the ori

[2] #16 6.RP.A.3 Use the $2{:}1$ ratios to locate $F$ and $D$. Side $AE$ has length $3$, and $AF

[3] #7 6.G.A.1 Break the square into $\triangle BFD$ plus three right triangles in the corners

[4] #7 6.G.A.1 Compute the three corner right-triangle areas. Each has its right angle at a squ

[5] #7 6.RP.A.1 Subtract to get the area of $\triangle BFD$, then divide by the square's area to

Review

Reasonableness: Quick size check: the three corner triangles together have area $3 + 3 + \tfrac{1}{2} = \tfrac{13}{2}$, which is a bit more than half of the square's area $9$. That leaves $\tfrac{5}{2}$ for $\triangle BFD$, a little less than half the square. The ratio $\tfrac{5}{18} \approx 0.278$ matches that: a touch over a quarter. Among the choices, $\tfrac{1}{6}\approx 0.17$ is too small, $\tfrac{1}{3}\approx 0.33$ is too big, and only $\tfrac{5}{18}$ lands in the right window.

Alternative: Tool #3 (Use Coordinates / Shoelace): with $B=(3,3)$, $F=(0,1)$, $D=(1,0)$, the shoelace formula gives $[\triangle BFD] = \tfrac{1}{2}\,|3(1-0) + 0(0-3) + 1(3-1)| = \tfrac{1}{2}\,|3 + 0 + 2| = \tfrac{5}{2}$. Dividing by $9$ again gives $\tfrac{5}{18}$, confirming (C).

CCSS standards used (min grade 6)

6.G.A.3 Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find side lengths (Placing square $ABCE$ on a coordinate grid with $E$ at the origin and reading off $F=(0,1)$, $D=(1,0)$ from the $2{:}1$ side splits.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Turning the $2{:}1$ ratios on sides $AE$ and $CE$ into integer lengths after scaling the square to side $3$.)
6.G.A.1 Find the area of right triangles and other polygons by composing into rectangles or decomposing into triangles (Decomposing square $ABCE$ into $\triangle BFD$ plus three corner right triangles and computing each corner area as $\tfrac{1}{2}\cdot\text{leg}\cdot\text{leg}$.)
6.RP.A.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship (Forming and simplifying the final area ratio $\dfrac{5/2}{9} = \dfrac{5}{18}$.)

⭐ Put the square on graph paper at side $3$, slice off the three corner triangles, and what is left is $\triangle BFD$ — a clean Grade 6 area-decomposition trick that turns a tricky ratio into simple subtraction.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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