AMC 8 · 2022 · #18
Grade 6 geometry-2dProblem
The midpoints of the four sides of a rectangle are and What is the
area of the rectangle?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: A rectangle has its four side-midpoints located at $(-3,0)$, $(2,0)$, $(5,4)$, and $(0,4)$. Using only these four midpoints (not the rectangle's corners), find the area of the rectangle.
Givens: Four midpoints (one per side): $(-3,0)$, $(2,0)$, $(5,4)$, $(0,4)$; These are midpoints of the four sides of a rectangle (the rectangle itself is not drawn); Answer choices: (A) $20$, (B) $25$, (C) $40$, (D) $50$, (E) $80$
Unknowns: The area of the original rectangle
Understand
Restated: A rectangle has its four side-midpoints located at $(-3,0)$, $(2,0)$, $(5,4)$, and $(0,4)$. Using only these four midpoints (not the rectangle's corners), find the area of the rectangle.
Givens: Four midpoints (one per side): $(-3,0)$, $(2,0)$, $(5,4)$, $(0,4)$; These are midpoints of the four sides of a rectangle (the rectangle itself is not drawn); Answer choices: (A) $20$, (B) $25$, (C) $40$, (D) $50$, (E) $80$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem
The problem is purely spatial — four points on a grid — so Tool #1 (Draw a Diagram) is the natural first move: plot the four midpoints on graph paper and the shape jumps out. The picture turns the puzzle into two clean subproblems (Tool #7): first find the area of the easy inner parallelogram formed by the midpoints, then relate it to the rectangle's area. To discover that ratio without memorizing a theorem, Tool #9 (Solve an Easier Related Problem) is perfect — try a simple axis-aligned rectangle whose midpoints are easy to write down, see that its midpoint-rhombus has exactly half the rectangle's area, and reuse that ratio here.
Execute — Answer: C
5.G.A.2 Step 1 - Plot the four midpoints on graph paper and connect them in order: $A=(-3,0) \to B=(2,0) \to C=(5,4) \to D=(0,4) \to A$.
- Two sides ($AB$ and $DC$) lie on horizontal lines ($y=0$ and $y=4$), and the picture shows a tilted parallelogram sitting inside the rectangle.
💡 Plotting four ordered pairs on a coordinate grid is exactly the Grade 5 coordinate-graphing skill.
6.G.A.3 Step 2 - Find the area of the inner parallelogram $ABCD$.
- Side $AB$ is horizontal at $y=0$ with length $|2-(-3)|=5$, and the opposite side $DC$ is horizontal at $y=4$, so the perpendicular height between them is $4$.
- Area $= \text{base} \times \text{height}$.
💡 Subproblem #1: use coordinates to read off the parallelogram's base and height directly from the plot — a Grade 6 coordinate-polygon move.
6.G.A.1 Step 3 - Discover the rectangle-to-midpoint-parallelogram ratio with an easier related problem.
- Take a simple $6 \times 4$ rectangle whose corners are $(0,0),(6,0),(6,4),(0,4)$.
- Its side midpoints are $(3,0),(6,2),(3,4),(0,2)$, forming a rhombus with diagonals of length $6$ and $4$.
- The rhombus area is $\tfrac{1}{2}\times 6 \times 4 = 12$, which is exactly half of the rectangle's area $6 \times 4 = 24$.
- So midpoint-shape area $= \tfrac{1}{2} \times$ rectangle area, no matter which rectangle.
💡 Trying a simple axis-aligned rectangle shows the $\tfrac{1}{2}$ ratio with no theorem needed — a Grade 6 composing/decomposing argument.
6.G.A.1 Step 4 - Apply the ratio to our problem.
- Since the midpoint parallelogram has area $20$ and equals $\tfrac{1}{2}$ of the rectangle's area, the rectangle's area must be twice $20$.
💡 Subproblem #2: combine the inner area with the rectangle-to-midpoint ratio to get the answer — Grade 6 area composition.
5.G.A.2 Plot the four midpoints on graph paper and connect them in order: $A=(-3,0) \to 6.G.A.3 Find the area of the inner parallelogram $ABCD$. Side $AB$ is horizontal at $y=0 6.G.A.1 Discover the rectangle-to-midpoint-parallelogram ratio with an easier related pr 6.G.A.1 Apply the ratio to our problem. Since the midpoint parallelogram has area $20$ a Review
Reasonableness: Look at the picture: the four midpoints stretch from $x=-3$ to $x=5$ (width $8$) and from $y=0$ to $y=4$ (height $4$), so they fit inside a bounding box of area $8 \times 4 = 32$. The actual rectangle is tilted, but its area should be a bit larger than the inner parallelogram ($20$) and on the same order as the bounding box. $40$ sits comfortably between those — and it matches choice (C). Choices (A) $20$ would mean the rectangle equals the inner parallelogram (impossible), and (E) $80$ would mean the rectangle is four times bigger, which would burst past the bounding box.
Alternative: Tool #13 (Convert to Algebra): the rectangle's sides are parallel to vectors $\vec{AB}=(5,0)$ and $\vec{BC}=(3,4)$ scaled by $2$ (each side of the rectangle is twice the length of the corresponding midpoint-rhombus diagonal segment direction). Computing the rectangle's actual side lengths $|2\vec{u}|$ and $|2\vec{v}|$ for perpendicular $\vec{u},\vec{v}$ recovers the same area $40$, but this route relies on vectors and is heavier than the diagram-plus-easier-case path.
CCSS standards used (min grade 6)
5.G.A.2Represent real-world and mathematical problems by graphing points (Plotting the four given midpoints $(-3,0),(2,0),(5,4),(0,4)$ on the coordinate plane to see the inner parallelogram.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Reading the parallelogram's base ($5$ units, horizontal) and height ($4$ units, vertical) directly from the coordinates to compute its area $5 \times 4 = 20$.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Using an easier $6 \times 4$ rectangle to discover that the midpoint-shape always has half the rectangle's area, then doubling $20$ to get $40$.)
⭐ This AMC 8 problem only needs Grade 6 coordinate-geometry area skills you already know — plot the points, find the inner shape's area, double it!
⭐ This AMC 8 problem only needs Grade 6 coordinate-geometry area skills you already know — plot the points, find the inner shape's area, double it!
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