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

A 3 × 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.

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.

How to solve

Strategy Change Focus / Count the Complement

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 △ CGB) - (area of bottom 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.

1STEP 1

Drop the rectangle onto a grid so EB and FC become honest lines; call their crossing inside trapezoid EFCB point G.

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

2STEP 2

The wings live inside trapezoid EFCB, whose parallel bases EF = 3 and CB = 1 with height 4 give area 8.

Area(EFCB) =

\frac{1}{2}

(3 + 1)(4) =

\frac{1}{2}

(4)(4) = 8

3STEP 3

Since CB ∥ EF, triangles CGB and EGF are similar with base ratio 1 : 3, so the heights split as 1 and 3 (they sum to 4).

h_top + h_bot = 4, h_bot = 3 h_top → h_top = 1, h_bot = 3

4STEP 4

The unshaded triangles come out to $\frac{1}{2}$ (top: base 1, height 1) and $\frac{9}{2}$ (bottom: base 3, height 3).

Area(△ CGB) =

\frac{1}{2}

(1)(1) =

\frac{1}{2}

, Area(△ EGF) =

\frac{1}{2}

(3)(3) =

\frac{9}{2}

5STEP 5

Subtract both triangles from the trapezoid: 8 - $\frac{1}{2}$ - $\frac{9}{2}$ = 3, the two bat wings.

Shaded = 8 -

\frac{1}{2}

\frac{9}{2}

= 8 -

\frac{10}{2}

= 8 - 5 = 3 → (C)

Answer

The trapezoid has area 8 and the inner unshaded triangles take up

\frac{1}{2}

\frac{9}{2}

= 5, leaving 3 for the wings. That's

\frac{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 × 4 = 12, and 3 is

\frac{1}{4}

of 12, a clean fraction consistent with the evenly-spaced setup. Answer (C) = 3 is the only choice that matches.

💡Key takeaway

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.

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