AMC 8 · 2015 · #21
Grade 8 geometry-2dProblem
In the given figure, hexagon ABCDEF is equiangular, ABJI and FEHG are squares with areas 18 and 32 respectively, △JBK is equilateral and FE=BC. What is the area of △KBC?
Pick an answer.
AMC 8 2015 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.
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
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) $32$
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
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.
💡 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.
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 $\angle ABC = 120^\circ$.
💡 The interior-angle formula $(n-2) \cdot 180^\circ / n$ is the same idea used to find angles in any equiangular polygon.
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 $\angle KBJ = 60^\circ$.
💡 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.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 $\angle KBC$.
💡 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 $B$.
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 $= \tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$.
💡 The two $\sqrt{2}$ factors multiply to $2$, killing the radical. The triangle-area formula then gives a clean integer.
8.EE.A.2 Get the two side lengths from the square areas. A square's side is the square ro 8.G.A.5 Find the interior angle of the equiangular hexagon at $B$. The interior angles o 4.G.A.2 Read off the other two angles at $B$ from the known shapes. Square $ABJI$ has $\ 7.G.B.5 The four angles $\angle ABC$, $\angle JBA$, $\angle KBJ$, $\angle KBC$ fit aroun 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.2Classify 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.1Find 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.5Use 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.5Use 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.2Use 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.
⭐ 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.
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