AMC 10 · 2014 · #16
Grade 8 geometry-2dIn rectangle ABCD, AB=1, BC=2, and points E, F, and G are midpoints of BC, CD, and AD, respectively. Point H is the midpoint of GE. What is the area of the shaded region?
Pick an answer.
AMC 10 2014 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: A rectangle $ABCD$ has short side $AB=1$ and long side $BC=2$. Points $E$, $F$, $G$ are the midpoints of $\overline{BC}$, $\overline{CD}$, $\overline{AD}$, and $H$ is the midpoint of $\overline{GE}$. Segments $\overline{GE}$, the two-part path $A\text{–}F\text{–}B$, and the two-part path $D\text{–}H\text{–}C$ are drawn. Find the area of the small shaded kite where these segments cross near the middle.
Givens: Rectangle $ABCD$ with $AB=1$ and $BC=2$; $E$, $F$, $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, $\overline{AD}$ respectively; $H$ is the midpoint of $\overline{GE}$; The drawn segments are $\overline{GE}$, $\overline{AF}$, $\overline{FB}$, $\overline{DH}$, $\overline{HC}$; The shaded region is the kite bounded by $\overline{AF}, \overline{FB}, \overline{DH}, \overline{HC}$, with $H$ as its top corner and $F$ as its bottom corner; Answer choices: (A) $\dfrac1{12}$, (B) $\dfrac{\sqrt3}{18}$, (C) $\dfrac{\sqrt2}{12}$, (D) $\dfrac{\sqrt3}{12}$, (E) $\dfrac16$
Unknowns: The area of the shaded kite
Understand
Restated: A rectangle $ABCD$ has short side $AB=1$ and long side $BC=2$. Points $E$, $F$, $G$ are the midpoints of $\overline{BC}$, $\overline{CD}$, $\overline{AD}$, and $H$ is the midpoint of $\overline{GE}$. Segments $\overline{GE}$, the two-part path $A\text{–}F\text{–}B$, and the two-part path $D\text{–}H\text{–}C$ are drawn. Find the area of the small shaded kite where these segments cross near the middle.
Givens: Rectangle $ABCD$ with $AB=1$ and $BC=2$; $E$, $F$, $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, $\overline{AD}$ respectively; $H$ is the midpoint of $\overline{GE}$; The drawn segments are $\overline{GE}$, $\overline{AF}$, $\overline{FB}$, $\overline{DH}$, $\overline{HC}$; The shaded region is the kite bounded by $\overline{AF}, \overline{FB}, \overline{DH}, \overline{HC}$, with $H$ as its top corner and $F$ as its bottom corner; Answer choices: (A) $\dfrac1{12}$, (B) $\dfrac{\sqrt3}{18}$, (C) $\dfrac{\sqrt2}{12}$, (D) $\dfrac{\sqrt3}{12}$, (E) $\dfrac16$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The figure pins down every point by exact midpoints, so the cleanest move is Tool #1 (Draw a Diagram) upgraded to a coordinate grid: drop the rectangle onto axes and every corner becomes a simple fraction. Once coordinates exist, Tool #4 (Introduce a Variable) turns each border of the kite into a line equation $y=mx+b$, and the two side corners are just where pairs of those lines meet. Tool #7 (Identify Subproblems) then splits the finish into three small steps — find the crossing points, read the two diagonals, multiply — because the shaded shape is a kite whose diagonals happen to be one vertical and one horizontal segment. No circle, no trig, no similar-triangle chase is needed; the grid does the heavy lifting.
Execute — Answer: E
6.G.A.3 Step 1 Put the figure on coordinates
- Place $D$ at the origin with the long side going up.
- Since $AB=1$ and $BC=2$, set $D=(0,0)$, $C=(1,0)$, $B=(1,2)$, $A=(0,2)$.
- Then the midpoints are $E=(1,1)$ (of $\overline{BC}$), $F=(\tfrac12,0)$ (of $\overline{CD}$), $G=(0,1)$ (of $\overline{AD}$), and $H=(\tfrac12,1)$ (the midpoint of $\overline{GE}$).
- The shaded kite's top corner is $H$ and its bottom corner is $F$.
💡 A rectangle has perpendicular sides, so its corners and midpoints land on tidy grid fractions with zero effort.
8.EE.B.6 Step 2 Write the four border lines
- The kite is walled in by four segments.
- Using slope $=\dfrac{\text{rise}}{\text{run}}$ and the $y$-intercept for each: $\overline{AF}$ from $(0,2)$ to $(\tfrac12,0)$ has slope $-4$; $\overline{DH}$ from $(0,0)$ to $(\tfrac12,1)$ has slope $2$; $\overline{HC}$ from $(\tfrac12,1)$ to $(1,0)$ has slope $-2$; $\overline{FB}$ from $(\tfrac12,0)$ to $(1,2)$ has slope $4$.
💡 Two known points fix a line; slope-intercept form makes each wall ready to intersect.
8.EE.C.8 Step 3 Find the two side corners
- The left corner $X$ is where $\overline{AF}$ meets $\overline{DH}$; the right corner $Y$ is where $\overline{HC}$ meets $\overline{FB}$.
- Set each pair of equations equal and solve the little system.
💡 A corner is the single point sitting on both crossing segments, so make the two line equations agree.
6.NS.C.8 Step 4 Read off the two diagonals
- The kite's four corners are $H=(\tfrac12,1)$, $X=(\tfrac13,\tfrac23)$, $F=(\tfrac12,0)$, $Y=(\tfrac23,\tfrac23)$.
- Diagonal $\overline{HF}$ joins top to bottom; diagonal $\overline{XY}$ joins the two sides.
- $H$ and $F$ share $x=\tfrac12$, so $\overline{HF}$ is vertical with length $1-0=1$.
- $X$ and $Y$ share $y=\tfrac23$, so $\overline{XY}$ is horizontal with length $\tfrac23-\tfrac13=\tfrac13$.
💡 When two points share a coordinate, their distance is just the difference of the other coordinate — no square roots.
6.G.A.1 Step 5 Multiply the diagonals
- One diagonal is vertical and the other horizontal, so they cross at a right angle: the kite is orthodiagonal.
- Cutting it along its diagonals makes four right triangles whose areas sum to $\tfrac12 \cdot d_1 \cdot d_2$.
- So the shaded area is $\tfrac12\cdot 1\cdot\tfrac13=\tfrac16$, which is choice (E).
💡 For any quadrilateral whose diagonals cross at right angles, the area is half the product of those diagonals.
6.G.A.3 Place $D$ at the origin with the long side going up. Since $AB=1$ and $BC=2$, se 8.EE.B.6 The kite is walled in by four segments. Using slope $=\dfrac{\text{rise}}{\text{ 8.EE.C.8 The left corner $X$ is where $\overline{AF}$ meets $\overline{DH}$; the right co 6.NS.C.8 The kite's four corners are $H=(\tfrac12,1)$, $X=(\tfrac13,\tfrac23)$, $F=(\tfra 6.G.A.1 One diagonal is vertical and the other horizontal, so they cross at a right angl Review
Reasonableness: The shoelace formula on the four corners $H(\tfrac12,1), X(\tfrac13,\tfrac23), F(\tfrac12,0), Y(\tfrac23,\tfrac23)$ gives $\tfrac12|{-\tfrac13+\tfrac13+\tfrac13}|=\tfrac16$, matching the diagonal method. The whole rectangle has area $1\times2=2$, so the kite is $\tfrac1{12}$ of it — a believable little sliver squeezed near the center. The kite is also symmetric about the vertical line $x=\tfrac12$ (with $X$ and $Y$ mirror images), exactly as the picture looks.
Alternative: Skip the diagonal formula and use symmetry: the vertical diagonal $\overline{HF}$ has length $1$, and the left corner $X$ sits a horizontal distance $\tfrac12-\tfrac13=\tfrac16$ from the line $x=\tfrac12$. So triangle $HFX$ has area $\tfrac12\cdot 1\cdot\tfrac16=\tfrac1{12}$, and doubling for its mirror triangle $HFY$ gives $\tfrac16$. Same answer, no kite formula required.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing rectangle $ABCD$ on axes and locating every midpoint ($E,F,G,H$) as an exact grid fraction.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Turning each border segment of the kite into a slope-intercept line equation $y=mx+b$ from two known endpoints.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Finding the side corners $X$ and $Y$ as the intersection points of the pairs of border lines.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Measuring the two diagonals as coordinate differences, since $H,F$ share an $x$-value and $X,Y$ share a $y$-value.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Getting the kite's area by decomposing it along its perpendicular diagonals into right triangles, i.e. $\tfrac12 d_1 d_2$.)
⭐ Drop a right-angled figure onto a coordinate grid, turn each segment into a line equation, and a messy shaded region becomes a kite whose area is just half the product of its two diagonals.
⭐ Drop a right-angled figure onto a coordinate grid, turn each segment into a line equation, and a messy shaded region becomes a kite whose area is just half the product of its two diagonals.
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