AMC 8 · 2015 · #21

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$ . What is the area of $\triangle KBC$ ?

$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$ .

Pick an answer.

(A)

$6\sqrt{2}$

(B)

(C)

(D)

$9\sqrt{2}$

(E)

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Toolkit + CCSS Solution

Understand

Restated: 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 $\triangle JBK$ is built on side $JB$ of the first square. We are also told $FE = BC$. Find the area of $\triangle KBC$.

Givens: Hexagon $ABCDEF$ is equiangular; Square $ABJI$ has area $18$, so $AB = JB = \sqrt{18} = 3\sqrt{2}$; Square $FEHG$ has area $32$, so $FE = \sqrt{32} = 4\sqrt{2}$; $\triangle JBK$ is equilateral, so $BK = JB = 3\sqrt{2}$; $FE = BC$, so $BC = 4\sqrt{2}$; Answer choices: (A) $6\sqrt{2}$, (B) $9$, (C) $12$, (D) $9\sqrt{2}$, (E) $32$

Unknowns: The area of $\triangle KBC$

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #1 Draw a Diagram

The area of $\triangle KBC$ needs two ingredients: the two side lengths $BK$ and $BC$, and the angle $\angle KBC$ 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 $\angle KBC$ 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 $360^\circ$, so $\angle KBC$ is whatever is left after subtracting the other three.

Execute — Answer: C

#7 Identify Subproblems 8.EE.A.2 Step 1

Get the two side lengths from the square areas.
A square's side is the square root of its area, so $BC = FE = \sqrt{32}$ and $BK = JB = AB = \sqrt{18}$.
Simplify each radical by pulling out the largest perfect square factor.

$$BC = \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}, \quad BK = \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}$$

💡 A square of area $A$ has side $\sqrt{A}$ — a Grade 8 square-root fact. Pulling out the perfect square ($16$ from $32$, $9$ from $18$) leaves a clean $\sqrt{2}$ that will collapse later.

#7 Identify Subproblems 8.G.A.5 Step 2

Find the interior angle of the equiangular hexagon at $B$.
The interior angles of any hexagon sum to $(6-2) \cdot 180^\circ = 720^\circ$, and 'equiangular' means all six are equal, so each is $720^\circ / 6 = 120^\circ$.
Therefore $\angle ABC = 120^\circ$.

$$\angle ABC = \dfrac{(6-2) \cdot 180^\circ}{6} = \dfrac{720^\circ}{6} = 120^\circ$$

💡 The interior-angle formula $(n-2) \cdot 180^\circ / n$ is the same idea used to find angles in any equiangular polygon.

#1 Draw a Diagram 4.G.A.2 Step 3

Read off the other two angles at $B$ from the known shapes.
Square $ABJI$ has $\angle JBA = 90^\circ$.
Equilateral $\triangle JBK$ has $\angle KBJ = 60^\circ$.

$$\angle JBA = 90^\circ, \quad \angle KBJ = 60^\circ$$

💡 A square's corner is always a right angle; an equilateral triangle's corner is always $60^\circ$. These are Grade-4 shape facts.

#7 Identify Subproblems 7.G.B.5 Step 4

The four angles $\angle ABC$, $\angle JBA$, $\angle KBJ$, $\angle KBC$ fit around point $B$ with no overlap and no gap, so they sum to $360^\circ$.
Solve for $\angle KBC$.

$$120^\circ + 90^\circ + 60^\circ + \angle KBC = 360^\circ \;\Rightarrow\; \angle KBC = 360^\circ - 270^\circ = 90^\circ$$

💡 Angles around a point always add to $360^\circ$. Subtraction gives the missing angle — and here it lands on a clean $90^\circ$, so $\triangle KBC$ is a right triangle at $B$.

#7 Identify Subproblems 6.G.A.1 Step 5

Since $\angle KBC = 90^\circ$, the sides $BK$ and $BC$ are the two legs of a right triangle.
Use area $= \tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$.

$$\text{Area} = \tfrac{1}{2} \cdot BK \cdot BC = \tfrac{1}{2} \cdot 3\sqrt{2} \cdot 4\sqrt{2} = \tfrac{1}{2} \cdot 12 \cdot 2 = 12 \;\Rightarrow\; \textbf{(C)}$$

💡 The two $\sqrt{2}$ factors multiply to $2$, killing the radical. The triangle-area formula then gives a clean integer.

[1] #7 8.EE.A.2 Get the two side lengths from the square areas. A square's side is the square ro

[2] #7 8.G.A.5 Find the interior angle of the equiangular hexagon at $B$. The interior angles o

[3] #1 4.G.A.2 Read off the other two angles at $B$ from the known shapes. Square $ABJI$ has $\

[4] #7 7.G.B.5 The four angles $\angle ABC$, $\angle JBA$, $\angle KBJ$, $\angle KBC$ fit aroun

[5] #7 6.G.A.1 Since $\angle KBC = 90^\circ$, the sides $BK$ and $BC$ are the two legs of a rig

Review

Reasonableness: The square areas $18$ and $32$ have a ratio of about $1 : 1.78$, so the two legs $3\sqrt{2}$ and $4\sqrt{2}$ have ratio $3 : 4$. A right triangle with legs $3\sqrt{2}$ and $4\sqrt{2}$ has area $\tfrac{1}{2}(3\sqrt{2})(4\sqrt{2}) = 12$, sandwiched between the two square areas ($18$ and $32$) — which matches the visual expectation that $\triangle KBC$ is smaller than either square. Among the choices, $(C)\;12$ is the only one with no radical, which is consistent with $\sqrt{2} \cdot \sqrt{2} = 2$ tidying up the answer.

Alternative: Tool #3 (Eliminate Possibilities) on the choices: the two legs of $\triangle KBC$ both contain $\sqrt{2}$, so their product contains $\sqrt{2} \cdot \sqrt{2} = 2$ — a rational number. That immediately rules out $(A)\;6\sqrt{2}$ and $(D)\;9\sqrt{2}$. Choice $(E)\;32$ would require the triangle to be as big as the larger square, impossible. Between $(B)\;9$ and $(C)\;12$, the legs $3\sqrt{2}$ and $4\sqrt{2}$ already give an area of $\tfrac{1}{2} \cdot 12 \cdot 2 = 12$ assuming any angle of $90^\circ$ — and a quick check of the four angles around $B$ confirms that angle. Answer (C).

CCSS standards used (min grade 8)

4.G.A.2 Classify two-dimensional figures based on the presence of parallel or perpendicular lines, or angles of a specified size (Reading off $\angle JBA = 90^\circ$ from the fact that $ABJI$ is a square, and $\angle KBJ = 60^\circ$ from the fact that $\triangle JBK$ is equilateral.)
6.G.A.1 Find the area of right triangles, other triangles, and special quadrilaterals (Computing the area of right triangle $\triangle KBC$ as $\tfrac{1}{2} \cdot BK \cdot BC = \tfrac{1}{2}(3\sqrt{2})(4\sqrt{2}) = 12$.)
7.G.B.5 Use facts about supplementary, complementary, vertical, and adjacent angles to solve problems (Using the fact that the four adjacent angles around point $B$ sum to $360^\circ$ to solve $120^\circ + 90^\circ + 60^\circ + \angle KBC = 360^\circ$ for $\angle KBC = 90^\circ$.)
8.G.A.5 Use informal arguments to establish facts about the angle sum of polygons (Deriving the interior angle of an equiangular hexagon as $(6-2) \cdot 180^\circ / 6 = 120^\circ$, so $\angle ABC = 120^\circ$.)
8.EE.A.2 Use square root symbols to represent solutions to equations of the form $x^2 = p$; evaluate square roots of small perfect squares (Recovering the side lengths $BC = \sqrt{32} = 4\sqrt{2}$ and $BK = \sqrt{18} = 3\sqrt{2}$ from the square areas $32$ and $18$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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