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Answer: \[ 2/3 sqrt3 \]

Thanks

- Thread starter Nich6ls
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- Thread starter
- #1

Answer: \[ 2/3 sqrt3 \]

Thanks

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- Jun 20, 2014

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- Jan 30, 2018

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If the length of a side of the original equilateral triangle is "s" then the altitude is $\sqrt{s^2- (s/2)^2}= \sqrt{s^2- s^2/4}= \sqrt{3s^2/4}= \frac{s\sqrt{3}}{2}$.

So the first equilateral triangle has perimeter 3s while the second has perimeter $\frac{3s\sqrt{3}}{2}$. The ratio of those is $\frac{3s}{\frac{3s\sqrt{3}}{2}}= 3s\frac{2}{3s\sqrt{3}}= \frac{2}{\sqrt{3}}= \frac{2\sqrt{3}}{3}$.

(What you wrote, "2/3sqrt(3)" would correctly be interpreted as $\frac{2}{3\sqrt{3}}$, which is wrong, but I suspect you meant "(2/3)sqrt(3)"or $\frac{2}{3}\sqrt{3}= \frac{2\sqrt{3}}{3}$, which is correct.)