AMC 10 · 2005 · #20
Grade 8 geometry-2dAn equiangular octagon has four sides of length 1 and four sides of length 2/2, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
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: An eight-sided figure has all of its angles equal. Four of its sides have length $1$ and the other four have length $\tfrac{\sqrt{2}}{2}$, and the two lengths take turns so that no side sits next to another of its own length. Find the area inside the figure.
Givens: The octagon is equiangular: all eight interior angles are equal; Four sides have length $1$ and four sides have length $\tfrac{\sqrt{2}}{2}$; The lengths alternate, so every side of length $1$ is flanked by two sides of length $\tfrac{\sqrt{2}}{2}$, and the reverse; Answer choices: (A) $\tfrac{7}{2}$, (B) $\tfrac{7\sqrt{2}}{2}$, (C) $\tfrac{5+4\sqrt{2}}{2}$, (D) $\tfrac{4+5\sqrt{2}}{2}$, (E) $7$
Unknowns: The area enclosed by the octagon
Understand
Restated: An eight-sided figure has all of its angles equal. Four of its sides have length $1$ and the other four have length $\tfrac{\sqrt{2}}{2}$, and the two lengths take turns so that no side sits next to another of its own length. Find the area inside the figure.
Givens: The octagon is equiangular: all eight interior angles are equal; Four sides have length $1$ and four sides have length $\tfrac{\sqrt{2}}{2}$; The lengths alternate, so every side of length $1$ is flanked by two sides of length $\tfrac{\sqrt{2}}{2}$, and the reverse; Answer choices: (A) $\tfrac{7}{2}$, (B) $\tfrac{7\sqrt{2}}{2}$, (C) $\tfrac{5+4\sqrt{2}}{2}$, (D) $\tfrac{4+5\sqrt{2}}{2}$, (E) $7$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems
A shape problem cries out for a picture, so Tool #1 (Draw a Diagram) leads. Once the octagon is drawn with every angle marked $135^\circ$, the four short sides tilt at exactly $45^\circ$. That tilt is the key: if I extend the four long sides until they meet, they trap the octagon inside a square, and the only thing sticking out beyond the octagon are four little corner triangles. So instead of measuring the awkward eight-sided region directly, I use Tool #16 (Change Focus / Count the Complement): area of octagon $=$ area of the square $-$ the four corners. Tool #7 (Identify Subproblems) then splits the job into two clean pieces I already know how to do — the area of a square and the area of a right triangle.
Execute — Answer: A
8.G.A.5 Step 1 Every angle is 135 degrees
- The interior angles of any octagon add to $(8-2)\cdot 180^\circ = 1080^\circ$.
- Since the octagon is equiangular, all eight angles are equal, so each one is $1080^\circ \div 8 = 135^\circ$.
- Draw the figure with a long side lying flat on the bottom; because each corner turns by $180^\circ-135^\circ=45^\circ$, the two short sides next to that bottom side rise at a $45^\circ$ slant.
💡 Equal angles in an octagon must each be $135^\circ$, which makes the short sides lean at a tidy $45^\circ$.
6.G.A.1 Step 2 Box the octagon in a square
- The four long sides come in two horizontal and two vertical pairs.
- Extend each long side in a straight line until it meets its neighbors; the four lines close up into a square that wraps snugly around the octagon.
- What lies inside the square but outside the octagon is exactly four small triangles, one snipped off at each corner of the square.
- So the octagon is the square with its four corners cut away.
💡 It is easier to measure a neat square and remove the corners than to measure the jagged octagon head-on.
8.G.B.7 Step 3 Each corner triangle is a 45-45-90 with legs 1/2
- A corner triangle is cut off by one short side of the octagon, which becomes its slanted hypotenuse of length $\tfrac{\sqrt{2}}{2}$.
- Its two legs run along the square's edges and meet at the square's right-angle corner; since the short side slants at $45^\circ$, the triangle is a $45^\circ$-$45^\circ$-$90^\circ$ triangle with two equal legs $\ell$.
- By the Pythagorean theorem, $\ell^2+\ell^2=\left(\tfrac{\sqrt{2}}{2}\right)^2=\tfrac12$, so $2\ell^2=\tfrac12$, giving $\ell^2=\tfrac14$ and $\ell=\tfrac12$.
💡 Equal legs plus the Pythagorean theorem turn the short side of $\tfrac{\sqrt2}{2}$ into a clean leg of $\tfrac12$.
6.G.A.1 Step 4 Find the square's area and one corner's area
- Look at the top edge of the square.
- It runs across one long side of the octagon, length $1$, plus one triangle leg at each end, each $\tfrac12$.
- So the square's side is $\tfrac12+1+\tfrac12=2$, and its area is $2\times 2 = 4$.
- Each corner triangle is a right triangle with legs $\tfrac12$ and $\tfrac12$, so its area is $\tfrac12\cdot\tfrac12\cdot\tfrac12=\tfrac18$.
💡 The square's side is one long side plus two half-unit legs, and each corner is a small right triangle.
6.G.A.1 Step 5 Subtract the four corners
- Remove all four corner triangles from the square.
- Their total area is $4\cdot\tfrac18=\tfrac12$, so the octagon's area is $4-\tfrac12=\tfrac72$.
- Matching this against the choices, $\tfrac72$ is choice (A), so the answer is (A).
💡 Whole square minus four equal corners leaves the octagon.
8.G.A.5 The interior angles of any octagon add to $(8-2)\cdot 180^\circ = 1080^\circ$. S 6.G.A.1 The four long sides come in two horizontal and two vertical pairs. Extend each l 8.G.B.7 A corner triangle is cut off by one short side of the octagon, which becomes its 6.G.A.1 Look at the top edge of the square. It runs across one long side of the octagon, 6.G.A.1 Remove all four corner triangles from the square. Their total area is $4\cdot\tf Review
Reasonableness: The bounding square has area $4$, and the octagon is the square with four small corners shaved off, so its area must be a little less than $4$ — and $\tfrac72=3.5$ is exactly that, which passes the sanity check. A second check: place the octagon on a grid with a bottom-left vertex at $(0,0)$ and walk the sides, landing on vertices $(1,0),(1.5,0.5),(1.5,1.5),(1,2),(0,2),(-0.5,1.5),(-0.5,0.5)$; the shoelace formula on these points gives $\tfrac12\cdot 7 = \tfrac72$, confirming the answer independently.
Alternative: Instead of removing corners from a big square, slice the octagon up. Because every angle is $135^\circ$, the octagon splits neatly into a central pattern of unit squares plus right triangles whose legs are $\tfrac12$; each such triangle has half the area of a $\tfrac12$-by-$\tfrac12$... more cleanly, the same $45^\circ$ tilt lets you draw horizontal and vertical cuts that carve the octagon into a $1\times1$ center square, four $1\times\tfrac12$ rectangles on the sides, and four corner triangles, whose areas add to $1+4\cdot\tfrac12+4\cdot\tfrac18=1+2+\tfrac12=\tfrac72$. Same answer, (A).
CCSS standards used (min grade 8)
8.G.A.5Use informal arguments to establish facts about the angle sum and exterior angles of polygons (Finding that each interior angle of the equiangular octagon is $135^\circ$, so the short sides slant at $45^\circ$.)6.G.A.1Find the area of triangles and other polygons by composing into rectangles or decomposing into triangles (Writing the octagon as a bounding square minus four corner triangles, and computing the square's and triangles' areas.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Turning the short side (hypotenuse $\tfrac{\sqrt{2}}{2}$) of a $45^\circ$-$45^\circ$-$90^\circ$ corner triangle into equal legs of length $\tfrac12$.)
⭐ Trap the octagon inside a square, then subtract the four little corner triangles: $4-\tfrac12=\tfrac72$.
⭐ Trap the octagon inside a square, then subtract the four little corner triangles: $4-\tfrac12=\tfrac72$.
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