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Four equilateral triangles are inscribed within each other a
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23 Jun 2013, 00:26
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Four equilateral triangles are inscribed within each other as pictured here. If the perimeter of the outer triangle is 24, what is the area of the innermost triangle?
A. \(\frac{1}{6}\)
B. \(\sqrt{3}/8\)
C. \(\frac{1}{3}\)
D. \(\sqrt{2}/4\)
E. \(\sqrt{3}/4\)
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Re: Four equilateral triangles are inscribed within each other a
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23 Jun 2013, 01:12
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Four equilateral triangles are inscribed within each other as pictured here. If the perimeter of the outer triangle is 24, what is the area of the innermost triangle?
A. \(\frac{1}{6}\)
B. \(\sqrt{3}/8\)
C. \(\frac{1}{3}\)
D. \(\sqrt{2}/4\)
E. \(\sqrt{3}/4\)
The length of a side of the first inscribed triangle is half of that of outer triangle:
Attachment:
Untitled2.png
Since the length of the side of bigger triangle is 24/3=8, then the length of a side of the first inscribed triangle is 8/2=4.
The same way: The length of a side of the second inscribed triangle is 4/2=2; The length of a side of the third inscribed triangle (the smallest one) is 2/2=1.
The area of an equilateral triangle is \(side^2*\frac{\sqrt{3}}{4}\). Therefore, the are of the smallest triangle is \(\frac{\sqrt{3}}{4}\).
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.