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I have an equation that breaks down a relationship between the side lengths of $\frac{1}{2}$ of an equilateral triangle. I there is a sub step that I do not understand, I will list all steps for completeness. The step that I do not understand is step 3 sub-step 2. How does $\sqrt{ s^2 - \frac{s^2}{4} }$ become $\large = \sqrt{\frac{3s^2}{4}}$?

Step 1 - Divide triangle into 2 halves:

This leaves us with $h$, $s$, and $\frac{s}{2}$.

Step 1

Step 2 - Use Pythagorean theorem to get side lengths:

$(\frac{s}{2})^2 + h^2 = s^2$

or

$h = \sqrt{s^2 - (\frac{s}{2})^2}$

Step 3 - Simplify right side of the equation:

$\sqrt{ s^2 - ( \frac{s}{2}^2 ) } = \sqrt{ s^2 - \frac{s^2}{4} }$

$\large = \sqrt{\frac{3s^2}{4}}$

$\large = \frac{\sqrt{3}}{2}s$

GettingItDone
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    There should still be a square root over that expression. – David P Dec 30 '21 at 17:32
  • Because $1 - \frac 14 = \frac 34$. – WhatsUp Dec 30 '21 at 17:34
  • @WhatsUp Okay understood. I didn't make that link and also the notation confused me. I would have anticipated $\frac{3}{4}s^2$ but these are equivalent aren't they. Thanks, if you want to move comment into answer I can close this. – GettingItDone Dec 30 '21 at 17:37

1 Answers1

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$\sqrt{s^2 - \dfrac{s^2}{4}}$

Multiply by $4$ and then divide by $4$:

$\sqrt{\dfrac{4 s^2}{4} - \dfrac{s^2}{4}}$

Fraction minus fraction:

$\sqrt{\dfrac{4 s^2 - s^2}{4}}$

Simplify:

$\sqrt{\dfrac{3 s^2}{4}}$

Square roots in top and bottom:

$\dfrac{\sqrt{3 s^2}}{\sqrt{4}}$

Simplify:

$\dfrac{s \sqrt{3}}{2}$

Tiago Cavalcante
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