AMC 10 · 2014 · #18
Grade 8 geometry-2dA square in the coordinate plane has vertices whose y-coordinates are 0, 1, 4, and 5. What is the area of the square?
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 square sits in the coordinate plane, tilted so its sides are not horizontal or vertical. Its four corners have $y$-coordinates $0$, $1$, $4$, and $5$. Find the area of the square.
Givens: The four vertices have $y$-coordinates $0$, $1$, $4$, and $5$; The four points form a square (all sides equal, all angles right); Answer choices: (A) $16$, (B) $17$, (C) $25$, (D) $26$, (E) $27$
Unknowns: The area of the square
Understand
Restated: A square sits in the coordinate plane, tilted so its sides are not horizontal or vertical. Its four corners have $y$-coordinates $0$, $1$, $4$, and $5$. Find the area of the square.
Givens: The four vertices have $y$-coordinates $0$, $1$, $4$, and $5$; The four points form a square (all sides equal, all angles right); Answer choices: (A) $16$, (B) $17$, (C) $25$, (D) $26$, (E) $27$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #1 Draw a Diagram, #7 Identify Subproblems
The $x$-coordinates are hidden, so track only how much each side rises. Tool #4 (Introduce a Variable) names the run and rise of one side; because the next side is that side rotated $90^\circ$, its run and rise are the same two numbers swapped. Tool #1 (Draw a Diagram) turns the four heights into a clear bottom corner, a top corner, and two side corners. Tool #7 (Identify Subproblems) splits the work into two clean pieces: first find the two rises, then feed them into the Pythagorean theorem for the area.
Execute — Answer: B
6.NS.C.8 Step 1 Sort the heights and place the corners
- Sort the four $y$-coordinates: $0<1<4<5$.
- In a tilted square the lowest corner sits at $y=0$ and the highest, its diagonal opposite, at $y=5$.
- The remaining two heights $1$ and $4$ belong to the two corners next to the bottom corner.
- Picture the square balanced on that bottom corner like a diamond.
💡 Sorting the heights tells you instantly which corner is lowest, which is highest, and which two are the sides.
8.G.A.1 Step 2 Name the rise of each side
- Start at the bottom corner.
- Walking along one side changes height by some amount $q$; walking along the other side (which is the first side turned a right angle) changes height by some amount $p$.
- Turning a segment $90^\circ$ swaps its horizontal and vertical changes, so the two side-corners sit at heights $p$ and $q$ above the bottom corner, and the far corner sits at height $p+q$.
💡 Rotating a side by a right angle just swaps how far it goes across and how far it goes up.
6.EE.B.7 Step 3 Match the rises to the given heights
- The three corners above the bottom one are at heights $p$, $q$, and $p+q$.
- These must be the leftover values $1$, $4$, and $5$.
- The largest, $5$, has to be the sum, so $p+q=5$, which forces the two single rises to be $\{p,q\}=\{1,4\}$ (indeed $1+4=5$).
💡 The tallest side corner is the diagonal one, so it must equal the two smaller rises added together.
8.G.B.7 Step 4 Use the right triangle to get the area
- One side of the square is the hypotenuse of a right triangle whose legs are its run and rise, $1$ and $4$.
- The Pythagorean theorem gives $s^2 = 1^2 + 4^2 = 1 + 16 = 17$.
- Since the area of a square is exactly the side length squared, the area is $s^2 = 17$.
- That is choice (B).
- The trap answers $16$ and $25$ come from squaring just one leg ($4^2$ or $5^2$) and forgetting the other.
💡 A tilted side is the hypotenuse of a run-and-rise right triangle, and area is that hypotenuse squared.
6.NS.C.8 Sort the four $y$-coordinates: $0<1<4<5$. In a tilted square the lowest corner s 8.G.A.1 Start at the bottom corner. Walking along one side changes height by some amount 6.EE.B.7 The three corners above the bottom one are at heights $p$, $q$, and $p+q$. These 8.G.B.7 One side of the square is the hypotenuse of a right triangle whose legs are its Review
Reasonableness: The total height of the square is $5$, so the side length is a bit less than $5$ (a tilted square is taller than its side). A side just under $5$ has area just under $25$, and $17$ fits that — its square root is about $4.12$. The tilt is real because $\{p,q\}=\{1,4\}$ are different, so the square genuinely leans. Choices $16$ and $25$ are exactly the one-leg mistakes ($4^2$ and $5^2$), and $17$ survives the full Pythagorean count.
Alternative: Assign coordinates directly: put the bottom corner at $(0,0)$ and take side vectors $(4,1)$ and $(-1,4)$, which are perpendicular and equal length. The corners land at $(0,0)$, $(4,1)$, $(-1,4)$, $(3,5)$ with $y$-values $0,1,4,5$ — matching the problem. The side length squared is $4^2+1^2=17$, so the area is $17$.
CCSS standards used (min grade 8)
6.NS.C.8Solve real-world problems by graphing points in all four quadrants (Reading the four vertices as points in the coordinate plane and ordering them by height to locate the bottom, top, and side corners.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Using the fact that turning a side $90^\circ$ swaps its run and rise, so the two neighboring corners rise by $p$ and $q$.)6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Matching the corner heights to $1$, $4$, $5$ and solving $p+q=5$ to get $\{p,q\}=\{1,4\}$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing the side length squared $s^2 = 1^2 + 4^2 = 17$, which is the area of the square.)
⭐ For a tilted square, follow the up-and-down rise of two sides that meet at a corner; those two rises are the legs of a right triangle, and the side squared (the area) is just leg squared plus leg squared.
⭐ For a tilted square, follow the up-and-down rise of two sides that meet at a corner; those two rises are the legs of a right triangle, and the side squared (the area) is just leg squared plus leg squared.
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