AMC 10 · 2005 · #8
Grade 8 geometry-2dIn the figure, the length of side AB of square ABCD is 50, E is between B and H, and BE=1. What is the area of the inner square EFGH?
Pick an answer.
AMC 10 2005 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 large square $ABCD$ has side length $\sqrt{50}$. Four congruent right triangles are folded inside, leaving a smaller tilted square $EFGH$ in the middle. Point $E$ sits on segment $BH$ with $BE = 1$. Find the area of the inner square $EFGH$.
Givens: Outer square $ABCD$ has side $AB = \sqrt{50}$; The four corner right triangles are congruent (the figure has $90^\circ$ rotational symmetry); $E$ lies between $B$ and $H$, and $BE = 1$; Answer choices: (A) $25$, (B) $32$, (C) $36$, (D) $40$, (E) $42$
Unknowns: The length of a side of the inner square $EFGH$; The area of the inner square $EFGH$
Understand
Restated: A large square $ABCD$ has side length $\sqrt{50}$. Four congruent right triangles are folded inside, leaving a smaller tilted square $EFGH$ in the middle. Point $E$ sits on segment $BH$ with $BE = 1$. Find the area of the inner square $EFGH$.
Givens: Outer square $ABCD$ has side $AB = \sqrt{50}$; The four corner right triangles are congruent (the figure has $90^\circ$ rotational symmetry); $E$ lies between $B$ and $H$, and $BE = 1$; Answer choices: (A) $25$, (B) $32$, (C) $36$, (D) $40$, (E) $42$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #4 Introduce a Variable
The inner square's side is not handed to us directly, so break the picture into pieces you can measure. Tool #1 shows the outer square is really the inner square plus four congruent right triangles, and each outer side $\sqrt{50}$ is a triangle's hypotenuse with short leg $BE=1$. Tool #7 then splits the job into two easy subproblems: first find the triangle's long leg with the Pythagorean theorem, then notice the long leg is exactly the inner-square side plus one more short leg. Tool #4 lets you call the long leg a name and write $1^2 + b^2 = 50$.
Execute — Answer: C
6.G.A.1 Step 1 Decompose the figure
- Look at one corner triangle, say $\triangle BEC$.
- It has a right angle at $E$, its hypotenuse is the full outer side $BC = \sqrt{50}$, and its short leg is $BE = 1$.
- Because the four triangles are congruent, every short leg is $1$ and every outer side is a hypotenuse of length $\sqrt{50}$.
💡 The big square is just the inner square with four identical right triangles stuck onto its edges.
8.G.B.7 Step 2 Find the long leg
- Name the long leg of the triangle $b = EC$.
- The right triangle $BEC$ has legs $1$ and $b$ and hypotenuse $\sqrt{50}$, so the Pythagorean theorem gives $1^2 + b^2 = (\sqrt{50})^2$.
- Solve for $b$ and take the square root.
💡 Knowing the slanted side and one leg pins down the other leg exactly.
6.G.A.1 Step 3 Get the inner-square side
- The long leg $EC = 7$ runs along the cevian from $C$ and is cut into two parts by the inner-square vertex $F$: the inner-square side $EF$, plus the piece $FC$.
- That piece $FC$ is the short leg of the neighboring congruent triangle, so $FC = 1$.
- Subtract to get the side of $EFGH$.
💡 Each long leg overshoots the inner square by exactly one short leg, so the inner side is the difference of the two legs.
8.EE.A.2 Step 4 Compute the area
The area of the inner square is the side squared.
💡 Side $6$ means area $36$, which is answer (C).
6.G.A.1 Look at one corner triangle, say $\triangle BEC$. It has a right angle at $E$, i 8.G.B.7 Name the long leg of the triangle $b = EC$. The right triangle $BEC$ has legs $1 6.G.A.1 The long leg $EC = 7$ runs along the cevian from $C$ and is cut into two parts b 8.EE.A.2 The area of the inner square is the side squared. Review
Reasonableness: Add the pieces back up: four right triangles of legs $1$ and $7$ have total area $4 \cdot \tfrac{1}{2}\cdot 1 \cdot 7 = 14$, and the inner square has area $36$, giving $14 + 36 = 50$ — exactly the outer square's area $(\sqrt{50})^2 = 50$. Everything fits, so $36$ (C) is right. The choice $25$ (A) would be the inner side squared if you forgot to subtract the short leg ($7-1$ mistaken for $5$), and $42$ (E) is a distractor with no clean derivation.
Alternative: Skip the inner side entirely (Tool #16, count the complement): the inner square is what's left after removing the four corner triangles from the outer square. Outer area $= 50$; four triangles $= 4\cdot\tfrac12\cdot1\cdot7 = 14$; so inner area $= 50 - 14 = 36$. Same answer, reached by subtraction instead of finding a side.
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding the triangle's long leg from the hypotenuse $\sqrt{50}$ and short leg $1$: $1^2 + b^2 = 50$ gives $b = 7$.)6.G.A.1Find the area of polygons by composing into rectangles or decomposing into triangles (Splitting the outer square into the inner square plus four congruent right triangles, and reading the inner side as long leg minus short leg.)8.EE.A.2Use square root symbols and evaluate square roots of small perfect squares (Evaluating $\sqrt{49} = 7$ for the long leg and squaring the side $6^2 = 36$ for the area.)
⭐ Cut the big square into an inner square plus four matching right triangles, use the Pythagorean theorem to get the long leg, and the inner side is just the long leg minus the short one.
⭐ Cut the big square into an inner square plus four matching right triangles, use the Pythagorean theorem to get the long leg, and the inner side is just the long leg minus the short one.
More like this
Same archetype — closest grade level first.
