AMC 10 · 2011 · #11
Grade 8 geometry-2dSquare EFGH has one vertex on each side of square ABCD. Point E is on AB with AE=7⋅EB. What is the ratio of the area of EFGH to the area of ABCD?
Pick an answer.
AMC 10 2011 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 tilted square $EFGH$ is inscribed inside a bigger square $ABCD$, with exactly one vertex of $EFGH$ on each side of $ABCD$. On side $AB$ the vertex $E$ splits the side so that the piece $AE$ is $7$ times the piece $EB$. Find how the area of the tilted square compares to the area of the big square.
Givens: $ABCD$ is a square; $EFGH$ is a square with one vertex on each side of $ABCD$; $E$ lies on side $AB$; $AE = 7 \cdot EB$; Answer choices: (A) $\tfrac{49}{64}$, (B) $\tfrac{25}{32}$, (C) $\tfrac{7}{8}$, (D) $\tfrac{5\sqrt{2}}{8}$, (E) $\tfrac{\sqrt{14}}{4}$
Unknowns: The ratio (area of $EFGH$) / (area of $ABCD$)
Understand
Restated: A tilted square $EFGH$ is inscribed inside a bigger square $ABCD$, with exactly one vertex of $EFGH$ on each side of $ABCD$. On side $AB$ the vertex $E$ splits the side so that the piece $AE$ is $7$ times the piece $EB$. Find how the area of the tilted square compares to the area of the big square.
Givens: $ABCD$ is a square; $EFGH$ is a square with one vertex on each side of $ABCD$; $E$ lies on side $AB$; $AE = 7 \cdot EB$; Answer choices: (A) $\tfrac{49}{64}$, (B) $\tfrac{25}{32}$, (C) $\tfrac{7}{8}$, (D) $\tfrac{5\sqrt{2}}{8}$, (E) $\tfrac{\sqrt{14}}{4}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #17 Visualize Spatial Relationships, #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
Tool #1 (Draw a Diagram) is the anchor: a labeled sketch of the tilted square inside the big square makes one corner right triangle appear, with its two legs sitting on one side of $ABCD$. Tool #4 (Introduce a Variable) picks a convenient side length ($8$) so that $AE = 7 \cdot EB$ turns into whole-number pieces $7$ and $1$. Tool #17 (Visualize Spatial Relationships) supplies the key move: a quarter-turn of the picture about the center carries $ABCD$ onto itself and $EFGH$ onto itself, so the neighboring side must be split in the SAME ratio — that fixes the second leg of the corner triangle. Then the Pythagorean theorem gives the tilted square's side, and Tool #3 (Eliminate Possibilities) confirms the resulting fraction against the five choices.
Execute — Answer: B
6.EE.B.7 Step 1 Pick a friendly side length
- Sizes are only compared as a ratio, so choose the side of $ABCD$ to be $8$.
- Then $AE + EB = 8$ with $AE = 7 \cdot EB$, which gives $8 \cdot EB = 8$, so $EB = 1$ and $AE = 7$.
💡 Choosing the side to be $8$ makes the $7:1$ split land on the whole numbers $7$ and $1$.
8.G.A.1 Step 2 Turn the picture a quarter-turn
- Rotate the whole figure $90^\circ$ about the center of $ABCD$.
- This rotation maps $ABCD$ onto itself and maps the inscribed square $EFGH$ onto itself, sending vertex $E$ (on $AB$) to vertex $F$ (on the next side $BC$).
- Because rotation preserves lengths, the piece $AE$ swings to the piece $BF$.
- So $BF = AE = 7$, and the remaining piece is $FC = EB = 1$.
💡 The inscribed square looks identical after a quarter-turn, so every side of the big square must be cut in the very same $7:1$ way.
8.G.B.7 Step 3 Find the tilted square's side
- Look at the corner triangle $EBF$ at vertex $B$.
- Its legs are $EB = 1$ and $BF = 7$, and the right angle sits at corner $B$ of the big square.
- The hypotenuse $EF$ is a side of the tilted square $EFGH$.
- Apply the Pythagorean theorem.
💡 Each side of the tilted square is the hypotenuse of a right triangle made from one long and one short piece of the big square's side.
6.EE.A.1 Step 4 Compare the two areas
- A square's area is its side squared.
- The tilted square's area is $EF^2 = 50$.
- The big square's area is $8^2 = 64$.
- Take the ratio and simplify by dividing top and bottom by $2$.
💡 Because area is side-squared, the side ratio's square is exactly the area ratio — and $EF^2$ is already the tilted area.
7.RP.A.2 Step 5 Match to a choice
- The ratio $\tfrac{25}{32}$ is exactly choice (B).
- Quick sanity on nearby choices: (A) $\tfrac{49}{64} \approx 0.766$, (C) $\tfrac{7}{8} = 0.875$, while $\tfrac{25}{32} \approx 0.781$ — only (B) matches.
- So the answer is (B).
💡 Turning the fraction into a decimal lets you line it up against the answer choices with no ambiguity.
6.EE.B.7 Sizes are only compared as a ratio, so choose the side of $ABCD$ to be $8$. Then 8.G.A.1 Rotate the whole figure $90^\circ$ about the center of $ABCD$. This rotation map 8.G.B.7 Look at the corner triangle $EBF$ at vertex $B$. Its legs are $EB = 1$ and $BF = 6.EE.A.1 A square's area is its side squared. The tilted square's area is $EF^2 = 50$. Th 7.RP.A.2 The ratio $\tfrac{25}{32}$ is exactly choice (B). Quick sanity on nearby choices Review
Reasonableness: The tilted square must be smaller than the big square but not by much, since $E$ sits near corner $B$ (only $1$ unit away) — the tilt is gentle. A gentle tilt should give a ratio a little below $1$, and $\tfrac{25}{32} \approx 0.78$ fits. Boundary check: if $E$ were at the midpoint the ratio would be $\tfrac{1}{2}$; if $E$ were at a corner the tilted square would coincide with the big one (ratio $1$). Our $7:1$ split is near a corner, so a ratio close to but under $1$ is exactly right.
Alternative: Coordinates (Tool #13, Convert to Algebra): place $A=(0,0)$, $B=(8,0)$, $C=(8,8)$, $D=(0,8)$. Then $E=(7,0)$. By the same $7:1$ split, $F=(8,7)$, $G=(1,8)$, $H=(0,1)$. Compute one side: $EF^2 = (8-7)^2 + (7-0)^2 = 1 + 49 = 50$, and the area ratio is $\tfrac{50}{64} = \tfrac{25}{32}$. Same answer without invoking symmetry explicitly.
CCSS standards used (min grade 8)
6.EE.B.7Solve real-world problems by writing and solving equations of the form px = q (Turning $AE + EB = 8$ with $AE = 7\,EB$ into $8\,EB = 8$ to get the pieces $EB = 1$ and $AE = 7$.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the squared side lengths $8^2 = 64$ and $EF^2 = 50$ to build the area ratio.)8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Using a length-preserving $90^\circ$ rotation to show the neighboring side is split in the same $7:1$ ratio, so $BF = AE = 7$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding the tilted square's side from the corner right triangle: $EF^2 = 1^2 + 7^2 = 50$.)7.RP.A.2Recognize and represent proportional relationships between quantities (Reading the area comparison as the ratio $\tfrac{50}{64} = \tfrac{25}{32}$ and matching it to a choice.)
⭐ Give the big square a handy side of $8$, split it $7$ and $1$, and notice a quarter-turn forces every side to split the same way — then one corner right triangle ($1$ and $7$) hands you the tilted square's side by the Pythagorean theorem, and $\tfrac{50}{64} = \tfrac{25}{32}$.
⭐ Give the big square a handy side of $8$, split it $7$ and $1$, and notice a quarter-turn forces every side to split the same way — then one corner right triangle ($1$ and $7$) hands you the tilted square's side by the Pythagorean theorem, and $\tfrac{50}{64} = \tfrac{25}{32}$.
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