AMC 10 · 2022 · #10
Grade 8 geometry-2dProblem
Daniel finds a rectangular index card and measures its diagonal to be 8 centimeters.
Daniel then cuts out equal squares of side 1 cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be 42 centimeters, as shown below. What is the area of the original index card?
Pick an answer.
AMC 10 2022 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: An index card is a rectangle with diagonal $8$ cm. From two opposite corners, $1$ cm × $1$ cm squares are removed. The two innermost vertices of those notches (one from each square) are $4\sqrt{2}$ cm apart. Find the area of the original rectangle.
Givens: Original rectangle has length $l$, width $w$, and diagonal $8$ cm; $1 \times 1$ squares removed at two opposite corners; Distance between the two inner notch-vertices is $4\sqrt{2}$ cm; Answer choices: (A) $14$, (B) $10\sqrt{2}$, (C) $16$, (D) $12\sqrt{2}$, (E) $18$
Unknowns: The area $lw$ of the original rectangle
Understand
Restated: An index card is a rectangle with diagonal $8$ cm. From two opposite corners, $1$ cm × $1$ cm squares are removed. The two innermost vertices of those notches (one from each square) are $4\sqrt{2}$ cm apart. Find the area of the original rectangle.
Givens: Original rectangle has length $l$, width $w$, and diagonal $8$ cm; $1 \times 1$ squares removed at two opposite corners; Distance between the two inner notch-vertices is $4\sqrt{2}$ cm; Answer choices: (A) $14$, (B) $10\sqrt{2}$, (C) $16$, (D) $12\sqrt{2}$, (E) $18$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #13 Convert to Algebra
Tool #1 (Draw a Diagram) — place the rectangle in coordinates with one cut corner at $(0,0)$ and the other at $(l,w)$. Mark the inner notch-vertices at $(1,1)$ and $(l-1, w-1)$. The picture makes both length facts directly visible: the diagonal of the rectangle is the segment from $(0,0)$ to $(l,w)$, and the segment between the inner notches is from $(1,1)$ to $(l-1, w-1)$ — same direction, just shortened by $\sqrt{2}$ at each end. Tool #7 (Identify Subproblems) — turn the two length facts into two equations using the Pythagorean theorem (or distance formula). Tool #13 (Convert to Algebra) — combine the two equations using the identity $(l+w)^2 = l^2 + w^2 + 2lw$ to get $lw$ directly, without ever solving for $l$ and $w$ individually.
Execute — Answer: E
6.G.A.3 Step 1 - Place the rectangle with corners $(0,0), (l, 0), (l, w), (0, w)$.
- The cut squares are at the opposite corners $(0,0)$ and $(l, w)$.
- Their inner notch-vertices are $(1,1)$ and $(l-1, w-1)$.
💡 Sticking the rectangle on a coordinate grid turns each labeled length into a Pythagorean computation.
8.G.B.7 Step 2 - The diagonal of the rectangle is the hypotenuse of the right triangle with legs $l$ and $w$.
- Apply the Pythagorean theorem with diagonal $= 8$.
💡 Length-width-diagonal forms a right triangle — the classic Pythagorean setup.
8.G.B.8 Step 3 - The segment between the two inner notch-vertices has horizontal change $(l-1)-1 = l-2$ and vertical change $(w-1)-1 = w-2$.
- Its length is $4\sqrt{2}$, so by the distance formula:
💡 Distance between two points in the plane is also a Pythagorean computation on coordinate differences.
8.EE.C.8 Step 4 - Expand the second equation: $(l-2)^2 + (w-2)^2 = l^2 - 4l + 4 + w^2 - 4w + 4 = (l^2+w^2) - 4(l+w) + 8 = 32$.
- Substitute the first equation $l^2+w^2 = 64$.
💡 Expanding and substituting collapses the two equations into a single linear fact about $l+w$.
7.EE.A.1 Step 5 - Use the identity $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw$ directly, without finding $l$ and $w$ separately.
- Plug in $l+w = 10$ and $l^2+w^2 = 64$.
💡 The square of a sum identity lets us trade $(l+w)^2$ and $l^2+w^2$ directly for the product $lw$ — the very thing we want (the area).
6.G.A.3 Place the rectangle with corners $(0,0), (l, 0), (l, w), (0, w)$. The cut square 8.G.B.7 The diagonal of the rectangle is the hypotenuse of the right triangle with legs 8.G.B.8 The segment between the two inner notch-vertices has horizontal change $(l-1)-1 8.EE.C.8 Expand the second equation: $(l-2)^2 + (w-2)^2 = l^2 - 4l + 4 + w^2 - 4w + 4 = ( 7.EE.A.1 Use the identity $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw$ directly, without f Review
Reasonableness: Plug back: with $l+w=10$ and $lw=18$, solve the quadratic $t^2 - 10t + 18 = 0$ to get $t = 5 \pm \sqrt{7}$, so $\{l, w\} = \{5+\sqrt{7},\;5-\sqrt{7}\}$. Check the diagonal: $l^2 + w^2 = (5+\sqrt{7})^2 + (5-\sqrt{7})^2 = (25+10\sqrt{7}+7) + (25-10\sqrt{7}+7) = 64$ ✓. Check the inner distance: $(l-2)^2 + (w-2)^2 = (3+\sqrt{7})^2 + (3-\sqrt{7})^2 = (9+6\sqrt{7}+7)+(9-6\sqrt{7}+7) = 32$ ✓. Area $= 18$ matches choice (E). The other choices fail: e.g. $14$ comes from forgetting the $+8$ when expanding; $16$ comes from a sign error.
Alternative: Tool #1 + Tool #8 (Analyze the Units) — observe that the diagonal of the inner segment lies along the same diagonal line of the rectangle. Its length is $l\sqrt{2} \cdot (\cos\theta)$-style ... actually a cleaner shortcut is: the inner segment $P_1 P_2$ is the original diagonal shrunk by $\sqrt{2}$ at each end (because each corner is pushed in by $(1,1)$, a vector of length $\sqrt{2}$ along the diagonal direction only when $l = w$). When $l \ne w$ the components of the corner-shift along and perpendicular to the diagonal differ, so this shortcut needs the algebra to make precise — the substitution-based path above is the cleanest.
CCSS standards used (min grade 8)
6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices (Placing the rectangle and notch-vertices on a coordinate grid: $(0,0), (l,0), (l,w), (0,w), (1,1), (l-1, w-1)$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Relating the rectangle's diagonal $8$ to its sides via $l^2 + w^2 = 64$.)8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Distance from $(1,1)$ to $(l-1, w-1)$: $(l-2)^2 + (w-2)^2 = 32$.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Subtracting $l^2+w^2=64$ from the expanded second equation to isolate $l+w = 10$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Using $(l+w)^2 = l^2 + 2lw + w^2$ to extract $lw = 18$ from $l+w$ and $l^2+w^2$.)
⭐ This AMC 10 problem only needs Grade 8 Pythagorean and algebra you already know — place the rectangle on a grid, write $l^2+w^2=64$ and $(l-2)^2+(w-2)^2=32$, get $l+w=10$, and use $(l+w)^2 = l^2+w^2+2lw$ to read off area $= lw = 18$.
⭐ This AMC 10 problem only needs Grade 8 Pythagorean and algebra you already know — place the rectangle on a grid, write $l^2+w^2=64$ and $(l-2)^2+(w-2)^2=32$, get $l+w=10$, and use $(l+w)^2 = l^2+w^2+2lw$ to read off area $= lw = 18$.