AMC 10 · 2002 · #17
Grade 8 geometry-2dA regular octagon ABCDEFGH has sides of length two. Find the area of △ADG.
Pick an answer.
AMC 10 2002 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 regular octagon $ABCDEFGH$ has every side equal to $2$. Its vertices are labeled in order around the shape. Connect vertices $A$, $D$, and $G$ to form a triangle, and find the area of $\triangle ADG$.
Givens: The octagon is regular: all eight sides have length $2$ and all eight interior angles are equal.; The vertices are labeled $A,B,C,D,E,F,G,H$ in order around the octagon.; The triangle uses vertices $A$ (1st), $D$ (4th), and $G$ (7th).
Unknowns: The area of $\triangle ADG$.
Understand
Restated: A regular octagon $ABCDEFGH$ has every side equal to $2$. Its vertices are labeled in order around the shape. Connect vertices $A$, $D$, and $G$ to form a triangle, and find the area of $\triangle ADG$.
Givens: The octagon is regular: all eight sides have length $2$ and all eight interior angles are equal.; The vertices are labeled $A,B,C,D,E,F,G,H$ in order around the octagon.; The triangle uses vertices $A$ (1st), $D$ (4th), and $G$ (7th).
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #17 Visualize Spatial Relationships
The triangle sits inside a shape whose exact corners are hard to picture from the labels alone, so the load-bearing move is tool #1 (Draw a Diagram): pin the octagon down in coordinates so every vertex has a known address. The clean way to get those coordinates is tool #17 (Visualize Spatial Relationships) — see the regular octagon as a square with four corners sliced off. Once $A$, $D$, and $G$ have coordinates, tool #7 (Identify Subproblems) turns the area into two easy pieces: a horizontal base and a vertical height.
Execute — Answer: C
8.G.B.7 Step 1 Slice the octagon from a square
- Picture a regular octagon as a big square with its four corners cut off.
- Each cut removes a right isosceles triangle whose slanted edge is one side of the octagon, so that slanted edge has length $2$.
- That edge is the hypotenuse, and the two equal legs run along the square's sides.
- If each leg is $\ell$, the Pythagorean theorem gives $\ell^2 + \ell^2 = 2^2$, so $2\ell^2 = 4$ and $\ell = \sqrt2$.
💡 A regular octagon is a square with equal right-triangle corners shaved off, and each shaved side is that triangle's hypotenuse.
6.G.A.3 Step 2 Put every vertex in coordinates
- The square's side is one octagon side plus two corner legs: $2 + 2\sqrt2$.
- Lay the octagon flat with its bottom edge horizontal.
- Going around, the vertices land at $A=(\sqrt2,\,0)$, $B=(\sqrt2+2,\,0)$, $C=(2\sqrt2+2,\,\sqrt2)$, $D=(2\sqrt2+2,\,\sqrt2+2)$, $E=(\sqrt2+2,\,2\sqrt2+2)$, $F=(\sqrt2,\,2\sqrt2+2)$, $G=(0,\,\sqrt2+2)$, and $H=(0,\,\sqrt2)$.
- The three we need are $A$, $D$, and $G$.
💡 Once the octagon is a square-minus-corners, each vertex address is just built from the side length $2$ and the leg $\sqrt2$.
6.G.A.3 Step 3 Read the base and height
- Look at $D=(2\sqrt2+2,\,\sqrt2+2)$ and $G=(0,\,\sqrt2+2)$: they share the same height $y=\sqrt2+2$, so segment $DG$ is horizontal.
- Use it as the base.
- Its length is the gap in the $x$-values, $(2\sqrt2+2)-0 = 2\sqrt2+2$.
- The height of the triangle is how far $A$ sits below that horizontal line: $A$ has $y=0$, so the drop is $\sqrt2+2$.
💡 Two points at the same height make a flat base, and the third point's vertical distance to it is the height.
6.G.A.1 Step 4 Multiply out the area
- A triangle's area is half the base times the height.
- So the area is $\tfrac12(2\sqrt2+2)(\sqrt2+2)$.
- Multiply the two factors: $(2\sqrt2+2)(\sqrt2+2) = 2\sqrt2\cdot\sqrt2 + 2\sqrt2\cdot2 + 2\cdot\sqrt2 + 2\cdot2 = 4 + 4\sqrt2 + 2\sqrt2 + 4 = 8 + 6\sqrt2$.
- Half of that is $4 + 3\sqrt2$.
- So the answer is (C).
💡 Half base times height still works even when the lengths carry a $\sqrt2$ — just expand carefully.
8.G.B.7 Picture a regular octagon as a big square with its four corners cut off. Each cu 6.G.A.3 The square's side is one octagon side plus two corner legs: $2 + 2\sqrt2$. Lay t 6.G.A.3 Look at $D=(2\sqrt2+2,\,\sqrt2+2)$ and $G=(0,\,\sqrt2+2)$: they share the same h 6.G.A.1 A triangle's area is half the base times the height. So the area is $\tfrac12(2\ Review
Reasonableness: Estimate with decimals: $\sqrt2\approx1.41$, so the base $2\sqrt2+2\approx4.83$ and the height $\sqrt2+2\approx3.41$, giving area $\approx\tfrac12(4.83)(3.41)\approx8.24$. The answer $4+3\sqrt2\approx8.24$ matches. For a sanity ceiling, the whole octagon has area $2(1+\sqrt2)\cdot2^2 = 8(1+\sqrt2)\approx19.3$, and the triangle's $8.24$ is a bit under half of it — believable for a big triangle spanning the octagon. Every length stayed positive and the result is one of the listed choices.
Alternative: Skip coordinates. Segment $DG$ cuts straight across the octagon; its length is one full side plus two sliced-corner legs, $2 + 2\sqrt2$. The perpendicular distance from $A$ to line $DG$ is one full side plus one corner leg, $2 + \sqrt2$. Then $\text{Area} = \tfrac12(2+2\sqrt2)(2+\sqrt2) = \tfrac12(8+6\sqrt2) = 4+3\sqrt2$, the same answer (C).
CCSS standards used (min grade 8)
8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Finding each sliced-corner leg as $\sqrt2$ from the octagon side of length $2$ as hypotenuse.)6.G.A.3Draw polygons in the coordinate plane given coordinates for the vertices; find side lengths (Placing the octagon in coordinates and reading off the base $DG$ and the height to $A$.)6.G.A.1Find the area of triangles and other polygons (Computing the triangle's area as half the base times the height.)
⭐ See the regular octagon as a square with its corners shaved off, drop it into coordinates, then any triangle's area is just half base times height.
⭐ See the regular octagon as a square with its corners shaved off, drop it into coordinates, then any triangle's area is just half base times height.
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