AMC 10 · 2006 · #20
Grade 8 geometry-2dIn rectangle ABCD, we have A=(6,−22), B=(2006,178), D=(8,y), for some integer y. What is the area of rectangle ABCD?
Pick an answer.
AMC 10 2006 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 three of its corners fixed on the coordinate plane: $A=(6,-22)$, $B=(2006,178)$, and $D=(8,y)$ for some whole number $y$. Corner $D$ sits directly across the $x$-value from $A$ but its height $y$ is unknown. Find the area of the rectangle.
Givens: $A=(6,-22)$ and $B=(2006,178)$ are two corners joined by side $AB$; $D=(8,y)$ is the corner joined to $A$ by side $AD$, with $y$ an integer; Answer choices: (A) $4000$, (B) $4040$, (C) $4400$, (D) $40{,}000$, (E) $40{,}400$
Unknowns: The area of rectangle $ABCD$; The height $y$ of corner $D$
Understand
Restated: A rectangle $ABCD$ has three of its corners fixed on the coordinate plane: $A=(6,-22)$, $B=(2006,178)$, and $D=(8,y)$ for some whole number $y$. Corner $D$ sits directly across the $x$-value from $A$ but its height $y$ is unknown. Find the area of the rectangle.
Givens: $A=(6,-22)$ and $B=(2006,178)$ are two corners joined by side $AB$; $D=(8,y)$ is the corner joined to $A$ by side $AD$, with $y$ an integer; Answer choices: (A) $4000$, (B) $4040$, (C) $4400$, (D) $40{,}000$, (E) $40{,}400$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems
The corners are given as coordinates, so Tool #1 (Draw a Diagram) turns the problem into a picture: two sides, $AB$ and $AD$, shoot out from corner $A$ and must meet at a right angle. Reading each side as a horizontal-and-vertical step makes the right angle usable. Tool #4 (Introduce a Variable) handles the unknown height $y$: the perpendicular condition gives one equation that pins it down. Tool #7 (Identify Subproblems) then splits the area into two clean pieces — measure the length of $AB$, measure the length of $AD$ — because area is just their product. A bonus of reading the sides as steps: the perpendicular side lengths multiply to a whole number even though each side is irrational.
Execute — Answer: E
6.NS.C.8 Step 1 Read each side as a horizontal-vertical step
- Sides $AB$ and $AD$ both start at corner $A=(6,-22)$.
- To reach $B=(2006,178)$ from $A$ you go right $2006-6=2000$ and up $178-(-22)=200$.
- To reach $D=(8,y)$ from $A$ you go right $8-6=2$ and up $y-(-22)=y+22$.
- So the step along $AB$ is $(\text{right }2000,\ \text{up }200)$ and the step along $AD$ is $(\text{right }2,\ \text{up }y+22)$.
💡 A side between two plotted points is just a right-and-up move, and its two numbers are the differences of the coordinates.
8.G.A.1 Step 2 Turn the corner 90 degrees to find y
- At corner $A$ the sides $AB$ and $AD$ meet at a right angle, so the step along $AD$ is the step along $AB$ turned a quarter turn.
- Turning a right-and-up step a quarter turn swaps its two numbers and flips one sign: turning $(2000,200)$ gives a direction shaped like $(200,-2000)$, i.e.
- the same as $(1,-10)$.
- Side $AD$ goes right $2$, so its step must be $2\cdot(1,-10)=(2,-20)$.
- Matching the vertical part: $y+22=-20$, so $y=-42$.
💡 A quarter turn trades a shape's horizontal and vertical reach and reverses one direction, so a perpendicular side reuses the same two numbers swapped.
8.G.B.8 Step 3 Measure both sides with the Pythagorean theorem
- Each side is the hypotenuse of a right triangle whose legs are its horizontal and vertical parts, so its length is $\sqrt{(\text{right})^2+(\text{up})^2}$.
- For $AB$ the legs are $2000$ and $200$: $\sqrt{2000^2+200^2}=\sqrt{4{,}040{,}000}=100\sqrt{404}$.
- For $AD$ the legs are $2$ and $20$: $\sqrt{2^2+20^2}=\sqrt{404}$.
💡 The distance between two points is the hypotenuse of the right triangle their horizontal and vertical gaps make.
6.G.A.1 Step 4 Multiply the two sides for the area
- The area of a rectangle is length times width, using two sides that meet at a corner — here $AB$ and $AD$.
- Multiplying, the two $\sqrt{404}$ factors combine: $\sqrt{404}\cdot\sqrt{404}=404$.
- So the area is $100\sqrt{404}\cdot\sqrt{404}=100\cdot404=40{,}400$.
- This is choice (E).
💡 Because the two sides share the factor $\sqrt{404}$, the messy square roots cancel into a clean whole number.
6.NS.C.8 Sides $AB$ and $AD$ both start at corner $A=(6,-22)$. To reach $B=(2006,178)$ fr 8.G.A.1 At corner $A$ the sides $AB$ and $AD$ meet at a right angle, so the step along $ 8.G.B.8 Each side is the hypotenuse of a right triangle whose legs are its horizontal an 6.G.A.1 The area of a rectangle is length times width, using two sides that meet at a co Review
Reasonableness: The horizontal gap alone from $A$ to $B$ is $2000$ and the vertical gap is $200$, so $AB$ is a bit over $2000$; side $AD$ is short, a bit over $20$. Their product should be a little above $2000\times20=40{,}000$, and $40{,}400$ sits exactly there — the two nearby choices (D) $40{,}000$ and (E) $40{,}400$ are the only ones in that range, and the tiny extra from the diagonal slant makes (E) the right one. It also checks that $y=-42$ is an integer, as the problem promised.
Alternative: You can skip finding $y$. Since area equals $AB\cdot AD$ and the two sides are perpendicular, form similar right triangles: $AB$ has legs $2000$ and $200$, and $AD$ (turned $90^\circ$) has its short leg $2$ playing the role that $200$ played, so $AD$ is $\tfrac{2}{200}=\tfrac1{100}$ as long as $AB$. Then $\text{Area}=AB\cdot AD=AB\cdot\tfrac{1}{100}AB=\tfrac{1}{100}AB^2=\tfrac{1}{100}(2000^2+200^2)=\tfrac{1}{100}(4{,}040{,}000)=40{,}400$, the same answer.
CCSS standards used (min grade 8)
6.NS.C.8Solve problems by graphing points in all four quadrants; find distances between points with the same first or second coordinate (Reading sides $AB$ and $AD$ as right-and-up steps by subtracting the coordinates of $A$ from $B$ and $D$.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Turning the step of side $AB$ a quarter turn to get the perpendicular direction of $AD$ and solving $y+22=-20$.)8.G.B.8Apply the Pythagorean Theorem to find the distance between two points in a coordinate system (Computing the side lengths $AB=100\sqrt{404}$ and $AD=\sqrt{404}$ from their horizontal and vertical legs.)6.G.A.1Find the area of right triangles, other triangles, special quadrilaterals, and polygons (Taking the area of the rectangle as the product of the two adjacent side lengths $AB$ and $AD$.)
⭐ Read each rectangle side as a right-and-up step; a square corner turns one step's numbers a quarter turn (swap and flip a sign), and multiplying the two side lengths lets the matching square roots cancel into a clean $40{,}400$.
⭐ Read each rectangle side as a right-and-up step; a square corner turns one step's numbers a quarter turn (swap and flip a sign), and multiplying the two side lengths lets the matching square roots cancel into a clean $40{,}400$.
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