AMC 8 · 2016 · #22

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

area-trianglessimilar-trianglescoordinate-geometry area-differenceidentify-subproblems ↑ Prerequisites: area-trianglessimilar-triangles

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$ . The area of the "bat wings" (shaded area) is

Pick an answer.

(A)

(B)

$2 \frac{1}{2}$

(C)

(D)

$3 \frac{1}{2}$

(E)

AMC 8 2016 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.

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Toolkit + CCSS Solution

Understand

Restated: A $3 \times 4$ rectangle $DEFA$ has its top side $DA$ split into three unit segments by points $C$ and $B$ (so $D, C, B, A$ sit at $x = 0, 1, 2, 3$). Two pairs of slanted lines are drawn: from $E$ to $B$ and from $F$ to $C$ (these cross to form an inner $\bowtie$ shape), and also from $E$ to $C$ and from $F$ to $B$ on the outside. The two black 'bat wing' triangles sit inside the trapezoid $EFCB$, on either side of where lines $EB$ and $FC$ cross. Find the total shaded area.

Givens: Rectangle $DEFA$ is $3$ wide and $4$ tall; $DC = CB = BA = 1$, so $D, C, B, A$ are evenly spaced across the top; The shaded 'bat wings' are two triangles formed by the segments $EB$, $FC$, $EC$, $FB$; Answer choices: (A) $2$, (B) $2\tfrac{1}{2}$, (C) $3$, (D) $3\tfrac{1}{2}$, (E) $5$

Unknowns: The combined area of the two shaded 'bat wing' triangles

Understand

Plan

Primary tool: #16 Change Focus / Count the Complement

Secondary: #1 Draw a Diagram, #7 Identify Subproblems

Adding up the two wing-shaped regions directly is awkward because their slanted boundaries make them hard to measure. Tool #16 (Count the Complement) flips the question: the wings live inside the trapezoid $EFCB$, and the rest of that trapezoid is just two clean triangles meeting at a single crossing point $G$. So shaded $=$ (area of trapezoid) $-$ (area of top triangle $\triangle CGB$) $-$ (area of bottom triangle $\triangle EGF$). Tool #1 (Draw a Diagram) puts the rectangle on coordinates so we can name the crossing point. Tool #7 (Subproblems) then splits the work into three independent pieces — the trapezoid area, the small top triangle, the large bottom triangle — each handled with a Grade-6 area formula.

Execute — Answer: C

#1 Draw a Diagram 5.G.A.1 Step 1

Place the rectangle on a coordinate grid: $E = (0, 0)$, $F = (3, 0)$, $A = (3, 4)$, $D = (0, 4)$, $C = (1, 4)$, $B = (2, 4)$.
The two crossing segments inside the trapezoid $EFCB$ are $EB$ (from $(0,0)$ to $(2,4)$) and $FC$ (from $(3,0)$ to $(1,4)$).
Call their crossing point $G$.

$$E(0,0),\; F(3,0),\; C(1,4),\; B(2,4)$$

💡 Coordinates turn 'slanted lines on a picture' into honest numbers so the area work later can't slip.

#7 Identify Subproblems 6.G.A.1 Step 2

Find the area of trapezoid $EFCB$.
The two horizontal sides are the parallel bases: bottom $EF = 3$, top $CB = 1$.
The vertical distance between them is the full rectangle height, $4$.
Apply the trapezoid area formula $\tfrac{1}{2}(b_1 + b_2) \cdot h$.

$$\text{Area}(EFCB) = \tfrac{1}{2}(3 + 1)(4) = \tfrac{1}{2}(4)(4) = 8$$

💡 The trapezoid is the whole container; everything we care about lives inside this $8$ square units.

#7 Identify Subproblems 8.G.A.5 Step 3

Locate the crossing point $G$ using similar triangles.
Because $CB \parallel EF$, the top triangle $\triangle CGB$ is similar to the bottom triangle $\triangle EGF$ (vertical angles at $G$, plus alternate interior angles on the parallel lines).
The base ratio is $CB : EF = 1 : 3$, so the height ratio is also $1 : 3$.
The two heights add to the full $4$, so the top height is $1$ and the bottom height is $3$.

$$h_{\text{top}} + h_{\text{bot}} = 4,\; h_{\text{bot}} = 3 h_{\text{top}} \;\Rightarrow\; h_{\text{top}} = 1,\; h_{\text{bot}} = 3$$

💡 Same shape but $3 \times$ bigger on the base means $3 \times$ taller too — so the crossing sits a quarter of the way down from the top.

#7 Identify Subproblems 6.G.A.1 Step 4

Compute the two unshaded triangle areas.
Top: base $CB = 1$, height $1$.
Bottom: base $EF = 3$, height $3$.

$$\text{Area}(\triangle CGB) = \tfrac{1}{2}(1)(1) = \tfrac{1}{2},\quad \text{Area}(\triangle EGF) = \tfrac{1}{2}(3)(3) = \tfrac{9}{2}$$

💡 The bottom triangle is wider and taller, so it eats most of the trapezoid; the tiny top triangle barely takes anything.

#16 Change Focus / Count the Complement 6.G.A.1 Step 5

Apply the complement: shaded wings $=$ trapezoid $-$ top triangle $-$ bottom triangle.

$$\text{Shaded} = 8 - \tfrac{1}{2} - \tfrac{9}{2} = 8 - \tfrac{10}{2} = 8 - 5 = 3 \;\Rightarrow\; \textbf{(C)}$$

💡 Subtracting the two clean triangles from the clean trapezoid leaves exactly the two bat wings — no slanted measuring needed.

[1] #1 5.G.A.1 Place the rectangle on a coordinate grid: $E = (0, 0)$, $F = (3, 0)$, $A = (3, 4

[2] #7 6.G.A.1 Find the area of trapezoid $EFCB$. The two horizontal sides are the parallel bas

[3] #7 8.G.A.5 Locate the crossing point $G$ using similar triangles. Because $CB \parallel EF$

[4] #7 6.G.A.1 Compute the two unshaded triangle areas. Top: base $CB = 1$, height $1$. Bottom:

[5] #16 6.G.A.1 Apply the complement: shaded wings $=$ trapezoid $-$ top triangle $-$ bottom tri

Review

Reasonableness: The trapezoid has area $8$ and the inner unshaded triangles take up $\tfrac{1}{2} + \tfrac{9}{2} = 5$, leaving $3$ for the wings. That's $\tfrac{3}{8}$ of the trapezoid — visually plausible from the figure, where the two wings together look like a bit under half of the trapezoid. Also note: the full rectangle is $3 \times 4 = 12$, and $3$ is $\tfrac{1}{4}$ of $12$, a clean fraction consistent with the evenly-spaced setup. Answer (C) $= 3$ is the only choice that matches.

Alternative: Tool #1 (Draw a Diagram) with coordinates handles it directly. Solve for $G$: line $EB$ is $y = 2x$ and line $FC$ is $y = -2x + 6$. Setting equal gives $x = \tfrac{3}{2},\; y = 3$, so $G = (1.5, 3)$. Then use the shoelace formula on each wing: left wing $E(0,0),\; C(1,4),\; G(1.5,3)$ has area $\tfrac{1}{2} |0(4-3) + 1(3-0) + 1.5(0-4)| = \tfrac{1}{2}|0 + 3 - 6| = \tfrac{3}{2}$. By the symmetry of the figure the right wing has the same area $\tfrac{3}{2}$, giving total $\tfrac{3}{2} + \tfrac{3}{2} = 3$. Same answer (C).

CCSS standards used (min grade 8)

5.G.A.1 Use a coordinate system; graph points in the first quadrant (Placing the rectangle at $E(0,0),\, F(3,0),\, C(1,4),\, B(2,4)$ so the slanted segments and the crossing point $G$ can be reasoned about with numbers, not just a picture.)
6.G.A.1 Find the area of right triangles, other triangles, and special quadrilaterals (Applying the trapezoid area formula for $EFCB$ ($\tfrac{1}{2}(3+1)(4) = 8$) and the triangle area formula for the two unshaded triangles ($\tfrac{1}{2}$ and $\tfrac{9}{2}$), then subtracting to get the shaded area $3$.)
8.G.A.5 Use informal arguments about angles and similarity (parallel lines cut by a transversal) (Recognizing $\triangle CGB \sim \triangle EGF$ from $CB \parallel EF$ (vertical angles plus alternate interior angles), so the $1 : 3$ base ratio forces the $1 : 3$ height ratio that splits the height $4$ into $1$ and $3$.)

⭐ Don't measure the weird wings directly — fill the trapezoid around them and subtract! Once you spot the $1 : 3$ similar triangles (Grade 8 idea), the heights split as $1$ and $3$ and the rest is one trapezoid area minus two triangle areas.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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