Problem 16 · 2016 Math Kangaroo
Hard
Geometry & Measurement
square-areapythagorean-triple
The square shown in the diagram has a perimeter of 4. The perimeter of the equilateral triangle is
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Answer: B — \(3 + \sqrt{3}\)
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Hint 1 of 2
The square has side 1; work out how far the triangle's side must reach past the square.
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Hint 2 of 2
The triangle's base extends beyond the square's foot by a segment set by the \(60^\circ\) slope, which has length \(\tfrac{1}{\sqrt3}\).
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Approach: read the triangle's side from the unit square
- The square has perimeter 4, so its side is 1.
- Where a slanted \(60^\circ\) side rises a height of 1 (the square's height), it runs sideways by \(\tfrac{1}{\sqrt3}\), the extra base length beyond the square.
- This makes the triangle's side \(1 + \tfrac{1}{\sqrt3}\), so its perimeter is \(3\left(1 + \tfrac{1}{\sqrt3}\right) = 3 + \sqrt3\).
- So the perimeter is \(3 + \sqrt3\) (B).
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