AMC 8 · 2015 · #21

Grade 8 geometry-2d

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

angle-sum-trianglearea-trianglesarea-rectanglesexponents identify-subproblemscasework ↑ Prerequisites: area-rectanglesarea-trianglesangle-sum-triangle

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Hexagon ABCDEF is equiangular (every interior angle equal). Square ABJI is attached to side AB and has area 18; square FEHG is attached to side FE and has area 32. A new equilateral triangle △ JBK is built on side JB of the first square. We are also told FE = BC. Find the area of △ KBC.

Pick an answer.

(A)

$6\sqrt{2}$

(B)

(C)

(D)

$9\sqrt{2}$

(E)

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The area of △ KBC needs two ingredients: the two side lengths BK and BC, and the angle ∠ KBC between them. Tool #7 (Identify Subproblems) splits the work into three clean sub-questions — (1) get BC and BK from the square areas, (2) get ∠ KBC from the four angles around point B, (3) plug into a triangle-area formula. Tool #1 (Draw a Diagram) is the supporting move: marking the four angles at B on the figure makes it visually obvious that they must fill up exactly 360°, so ∠ KBC is whatever is left after subtracting the other three.

1STEP 1

A square's side is the square root of its area, so BC = FE = √(32) = 4√(2) and BK = JB = √(18) = 3√(2).

BC = √(32) = √(16 · 2) = 4√(2), BK = √(18) = √(9 · 2) = 3√(2)

2STEP 2

A hexagon's interior angles sum to (6-2) · 180° = 720°, and equiangular splits that evenly, so ∠ ABC = 120°.

∠ ABC =

\frac{(6-2) · 180°}{6}

\frac{720°}{6}

= 120°

3STEP 3

Read the other two angles off the shapes: the square gives ∠ JBA = 90° and the equilateral triangle gives ∠ KBJ = 60°.

∠ JBA = 90°, ∠ KBJ = 60°

4STEP 4

The four angles at B fill 360°, so ∠ KBC = 360° - (120°+90°+60°) = 90° — a right angle at B.

120° + 90° + 60° + ∠ KBC = 360° → ∠ KBC = 360° - 270° = 90°

5STEP 5

With ∠ KBC = 90°, BK and BC are the legs, so area = $\frac{1}{2}$ · 3√(2) · 4√(2) = 12 → (C).

Area =

\frac{1}{2}

· BK · BC =

\frac{1}{2}

· 3√(2) · 4√(2) =

\frac{1}{2}

· 12 · 2 = 12 → (C)

Answer

The square areas 18 and 32 have a ratio of about 1 : 1.78, so the two legs 3√(2) and 4√(2) have ratio 3 : 4. A right triangle with legs 3√(2) and 4√(2) has area

\frac{1}{2}

(3√(2))(4√(2)) = 12, sandwiched between the two square areas (18 and 32) — which matches the visual expectation that △ KBC is smaller than either square. Among the choices, (C) 12 is the only one with no radical, which is consistent with √(2) · √(2) = 2 tidying up the answer.

💡Key takeaway

Split a tricky figure into pieces: side lengths from square areas, the missing angle from the 360° around a point, and a clean right-triangle area at the end.

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