AMC 10 · 2012 · #21
Grade 8 geometry-3dLet points A = (0,0,0), B = (1,0,0), C = (0,2,0), and D = (0,0,3). Points E, F, G, and H are midpoints of line segments BD, AB, AC, and DC respectively. What is the area of EFGH?
Pick an answer.
AMC 10 2012 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: Four points $A=(0,0,0)$, $B=(1,0,0)$, $C=(0,2,0)$, and $D=(0,0,3)$ sit in 3D space. $E$, $F$, $G$, $H$ are the midpoints of the segments $\overline{BD}$, $\overline{AB}$, $\overline{AC}$, $\overline{DC}$, in that order. Find the area of the quadrilateral $EFGH$.
Givens: The four corner points $A=(0,0,0)$, $B=(1,0,0)$, $C=(0,2,0)$, $D=(0,0,3)$.; $E$ is the midpoint of $\overline{BD}$, $F$ of $\overline{AB}$, $G$ of $\overline{AC}$, $H$ of $\overline{DC}$.; The vertices are joined in the order $E \to F \to G \to H$.
Unknowns: The area of quadrilateral $EFGH$.
Understand
Restated: Four points $A=(0,0,0)$, $B=(1,0,0)$, $C=(0,2,0)$, and $D=(0,0,3)$ sit in 3D space. $E$, $F$, $G$, $H$ are the midpoints of the segments $\overline{BD}$, $\overline{AB}$, $\overline{AC}$, $\overline{DC}$, in that order. Find the area of the quadrilateral $EFGH$.
Givens: The four corner points $A=(0,0,0)$, $B=(1,0,0)$, $C=(0,2,0)$, $D=(0,0,3)$.; $E$ is the midpoint of $\overline{BD}$, $F$ of $\overline{AB}$, $G$ of $\overline{AC}$, $H$ of $\overline{DC}$.; The vertices are joined in the order $E \to F \to G \to H$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
Put every point on the coordinate grid. Once the four midpoints have real coordinates, the shape of $EFGH$ stops being a mystery: I can break the job into small subproblems (find the corners, check the sides, then measure them) and use spatial reasoning to see which sides meet at a right angle.
Execute — Answer: C
6.NS.B.3 Step 1 Find the four midpoints
- A midpoint is the average of the two endpoints, taken one coordinate at a time.
- Averaging gives $E=\left(\tfrac12,0,\tfrac32\right)$, $F=\left(\tfrac12,0,0\right)$, $G=(0,1,0)$, and $H=\left(0,1,\tfrac32\right)$.
💡 The midpoint of a segment is just the average of its two endpoints, coordinate by coordinate.
6.NS.C.8 Step 2 Check opposite sides
- Walk each side by subtracting coordinates.
- Side $EF$ shifts by $\left(0,0,-\tfrac32\right)$ and side $GH$ shifts by $\left(0,0,\tfrac32\right)$: same line of travel, same length.
- Side $FG$ shifts by $\left(-\tfrac12,1,0\right)$ and side $HE$ shifts by $\left(\tfrac12,-1,0\right)$: again the same.
- So both pairs of opposite sides are equal and parallel, which makes $EFGH$ a parallelogram.
💡 If two sides shift by the exact same amounts, they must be equal in length and parallel.
4.G.A.1 Step 3 Find the right angle
- Look at where the sides point.
- Side $EF$ runs straight up and down the $z$-axis and has no $x$ or $y$ part.
- Side $FG$ lies completely flat in the $xy$-plane and has no $z$ part.
- A vertical segment and a flat segment meet at a square corner, so the parallelogram has a right angle.
- A parallelogram with a right angle is a rectangle.
💡 A segment going straight up meets a segment lying flat at a square corner.
8.G.B.8 Step 4 Measure the two sides
- Use the 3D distance formula (Pythagoras with three gaps) on the side-shift vectors.
- Side $EF$ has length $\tfrac32$.
- Side $FG$ has length $\sqrt{\tfrac14+1}=\tfrac{\sqrt5}{2}$.
💡 In 3D the straight-line distance is the square root of the squared gaps added together.
6.G.A.1 Step 5 Compute the area
- A rectangle's area is length times width.
- Multiplying the two side lengths gives $\tfrac32 \cdot \tfrac{\sqrt5}{2}=\tfrac{3\sqrt5}{4}$.
- That matches choice (C).
💡 A rectangle's area is just its length times its width.
6.NS.B.3 A midpoint is the average of the two endpoints, taken one coordinate at a time. 6.NS.C.8 Walk each side by subtracting coordinates. Side $EF$ shifts by $\left(0,0,-\tfra 4.G.A.1 Look at where the sides point. Side $EF$ runs straight up and down the $z$-axis 8.G.B.8 Use the 3D distance formula (Pythagoras with three gaps) on the side-shift vecto 6.G.A.1 A rectangle's area is length times width. Multiplying the two side lengths gives Review
Reasonableness: Numerically $\tfrac{3\sqrt5}{4}\approx 1.68$, which sits sensibly between sides of length $1.5$ and about $1.12$. The right-angle check ($EF\cdot FG=0$) and the equal opposite sides both confirm a genuine rectangle, so length times width is the correct area rule. Answer (C) holds.
Alternative: Skip the rectangle argument and use the cross product of two adjacent side vectors, whose magnitude is the parallelogram's area directly: $EF\times FG=\left(\tfrac32,\tfrac34,0\right)$, and $\left|EF\times FG\right|=\sqrt{\tfrac94+\tfrac{9}{16}}=\sqrt{\tfrac{45}{16}}=\tfrac{3\sqrt5}{4}$, the same answer.
CCSS standards used (min grade 8)
6.NS.B.3Fluently add, subtract, multiply, and divide multi-digit decimals (Averaging endpoint coordinates to locate each of the four midpoints.)6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Comparing coordinate shifts to show opposite sides are equal and parallel.)4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Identifying the perpendicular pair of sides that creates a right angle.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Measuring the two side lengths of the quadrilateral in 3D.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Computing the rectangle's area as length times width.)
⭐ Give every point coordinates, average them to find midpoints, and the hidden shape turns out to be a plain rectangle you can just measure.
⭐ Give every point coordinates, average them to find midpoints, and the hidden shape turns out to be a plain rectangle you can just measure.
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