Problem 3 · 2026 AMC 8
Medium
Geometry & Measurement
perimeterpythagorean-triplesquare-area
Haruki has a piece of wire that is 24 centimeters long. He wants to bend it to form each of the following shapes, one at a time.
- A regular hexagon with side length 5 cm.
- A square of area 36 cm2.
- A right triangle whose legs are 6 and 8 cm long.
Which of the shapes can Haruki make?
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Answer: D — Square and triangle only.
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Hint 1 of 2
The wire is one fixed length and can't stretch. So what single number must every makeable shape's outline equal?
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Hint 2 of 2
A shape works only if its perimeter equals 24 cm. Find each perimeter — and for the triangle, watch for a familiar right-triangle (the 6-8-10) so you don't have to compute a square root.
Show solution
Approach: the wire never stretches — only perimeter = 24 works
- A fixed loop of wire can only bend into a shape whose perimeter is exactly 24 cm. So the whole problem becomes: find each perimeter and check it against 24.
- Hexagon: 6 sides × 5 = 30 cm. Too long — no. (Notice you don't even need the others to rule this one out.)
- Square of area 36: the side is √36 = 6, so perimeter 4 × 6 = 24 cm. ✓
- Right triangle, legs 6 and 8: spot the 6-8-10 Pythagorean triple — the hypotenuse is 10, so perimeter 6 + 8 + 10 = 24 cm. ✓
- Only the square and the triangle fit, so the answer is Square and triangle only.
- You'll see it again: 3-4-5 and its scalings (6-8-10, 9-12-15…) are the most common right triangles on contests — recognizing one saves you the square-root work.
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