AMC 10 · 2008 · #17
Grade 7 geometry-2dAn equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point of the triangle?
Pick an answer.
AMC 10 2008 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 equilateral triangle has side length $6$. Look at every point that lies outside the triangle yet is within $3$ units of some point of the triangle. This is a band that hugs the outside of the triangle. Find the area of that band.
Givens: The triangle is equilateral with side length $6$; A point counts only if it is outside the triangle; A point counts only if its distance to the nearest point of the triangle is at most $3$; Answer choices: (A) $36+24\sqrt{3}$, (B) $54+9\pi$, (C) $54+18\sqrt{3}+6\pi$, (D) $\left(2\sqrt{3}+3\right)^2\pi$, (E) $9\left(\sqrt{3}+1\right)^2\pi$
Unknowns: The area of the outside band that stays within $3$ units of the triangle
Understand
Restated: An equilateral triangle has side length $6$. Look at every point that lies outside the triangle yet is within $3$ units of some point of the triangle. This is a band that hugs the outside of the triangle. Find the area of that band.
Givens: The triangle is equilateral with side length $6$; A point counts only if it is outside the triangle; A point counts only if its distance to the nearest point of the triangle is at most $3$; Answer choices: (A) $36+24\sqrt{3}$, (B) $54+9\pi$, (C) $54+18\sqrt{3}+6\pi$, (D) $\left(2\sqrt{3}+3\right)^2\pi$, (E) $9\left(\sqrt{3}+1\right)^2\pi$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #1 Draw a Diagram, #17 Visualize Spatial Relationships
The band has a curved, odd shape, so there is no single formula for it. Tool #7 (Identify Subproblems) says: cut the band into pieces that each have an easy formula. To see those pieces, Tool #1 (Draw a Diagram) sketches the band and shows it is made of flat strips along the sides and rounded wedges at the corners. Tool #17 (Visualize Spatial Relationships) helps read the exact turning angle of each corner wedge. Add the piece areas and the band's area falls out.
Execute — Answer: B
6.G.A.1 Step 1 Draw the band and split it into pieces
- Sketch the triangle and shade every outside point within $3$ units of it.
- Along each straight side the nearest point of the triangle is straight in, so the band there is a flat strip that reaches out exactly $3$ units: a rectangle.
- Near each corner the nearest point of the triangle is the corner itself, so all points within $3$ units of it form a rounded wedge (a slice of a circle of radius $3$).
- So the band is three rectangles plus three corner wedges.
💡 Every outside point is closest either to a flat side (giving a strip) or to a corner (giving a rounded wedge).
4.MD.A.3 Step 2 Add the three side rectangles
- Each rectangle runs the full length of a side, so it is $6$ long, and it reaches out $3$ units, so it is $3$ wide.
- Its area is $6 \times 3 = 18$.
- There are three equal sides, so the three rectangles together cover $3 \times 18 = 54$.
💡 A strip along a side is just a rectangle: length of the side times how far it reaches out.
4.MD.C.7 Step 3 Find the angle of one corner wedge
- At a corner, walk around the full turn of $360^\circ$.
- Part of that turn is blocked by the inside of the triangle: the corner angle of an equilateral triangle is $60^\circ$.
- Two more parts are the right-angle gaps where the wedge meets the two rectangles, since each rectangle sits square against its side: that is $90^\circ$ and $90^\circ$.
- What is left over for the rounded wedge is $360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ$.
💡 The angles around the corner point must add to a full turn, so subtracting the used-up parts leaves the wedge's angle.
7.G.B.4 Step 4 Combine the three wedges into one full circle
- Each of the three corners gives a $120^\circ$ wedge of a circle of radius $3$.
- Together they make $3 \times 120^\circ = 360^\circ$, which is one whole circle.
- A full circle of radius $3$ has area $\pi r^2 = \pi (3)^2 = 9\pi$.
- So all three corner wedges together add up to $9\pi$.
💡 Three $120^\circ$ slices snap together into one complete circle, so no partial-circle formula is needed.
6.G.A.1 Step 5 Add the pieces for the total area
- The band is the three rectangles plus the three wedges: $54 + 9\pi$.
- That matches answer choice (B).
💡 The whole band is just its flat strips plus its rounded corners added together.
6.G.A.1 Sketch the triangle and shade every outside point within $3$ units of it. Along 4.MD.A.3 Each rectangle runs the full length of a side, so it is $6$ long, and it reaches 4.MD.C.7 At a corner, walk around the full turn of $360^\circ$. Part of that turn is bloc 7.G.B.4 Each of the three corners gives a $120^\circ$ wedge of a circle of radius $3$. T 6.G.A.1 The band is the three rectangles plus the three wedges: $54 + 9\pi$. That matche Review
Reasonableness: The area splits cleanly into a plain number $54$ from the rectangles and a $\pi$-part $9\pi$ from the round corners, so the answer should look like $54 + (\text{something})\pi$. Only (B) $54+9\pi$ has that exact shape. Choice (A) has no $\pi$ at all, and (C) carries an extra $\sqrt{3}$ that no piece of this band produces, while (D) and (E) are a single $\pi$ term with no separate whole-number part. Numerically $54 + 9\pi \approx 82.3$, a sensible size for a $3$-wide band around a triangle of side $6$.
Alternative: Keep the wedges separate instead of merging them. Each corner wedge is $\tfrac{120}{360} = \tfrac{1}{3}$ of a circle of radius $3$, so its area is $\tfrac{1}{3}\,\pi(3)^2 = 3\pi$. Three of them give $3 \times 3\pi = 9\pi$, and adding the $54$ from the rectangles again gives $54 + 9\pi$, choice (B).
CCSS standards used (min grade 7)
6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Splitting the outside band into three rectangles plus three corner wedges, then adding the pieces to get the total area.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing each side strip as a rectangle of area $6 \times 3 = 18$, giving $54$ for all three.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems (Finding each corner wedge's angle as $360^\circ - 60^\circ - 90^\circ - 90^\circ = 120^\circ$.)7.G.B.4Know the formulas for area and circumference of a circle (Turning the three $120^\circ$ wedges into one full circle of radius $3$ with area $\pi(3)^2 = 9\pi$.)
⭐ The outside band is three flat strips ($54$) plus three corner slices that join into one full circle ($9\pi$), so the area is $54 + 9\pi$.
⭐ The outside band is three flat strips ($54$) plus three corner slices that join into one full circle ($9\pi$), so the area is $54 + 9\pi$.
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