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From an equilateral triangle $T$ where each side have a length of $L$. What is the area of $T$?

According to the Wikipedia page of equilateral triangles, the area is $$A=\dfrac{\sqrt{3}}{4}L^2$$

I am trying to solve this problem by using the Pythagorean theorem, as explained in this question, I can split the triangle in half to try and get the height.

Using the Pythagorean theorem, $$L^2=(\dfrac{L}{2})^2 + H^2$$

I can then isolate $H$ with :

$$H=\sqrt{L^2-(\dfrac{L}{2})^2}$$

Using the $A=\dfrac{1}{2}bh$ formula. I could then conclude with : $$A=\dfrac{L\sqrt{L^2-(\dfrac{L}{2})^2}}{2}$$

As said previously, the Wikipedia page shows something very different. What went wrong?

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  • $\begingroup$ Why do you think those are different? $\endgroup$ – Matthew Leingang Sep 4 '18 at 19:14
  • $\begingroup$ Factor the $L^2$ then pass it out. $\endgroup$ – Randall Sep 4 '18 at 19:15
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    $\begingroup$ $L^2-(L/2)^2=\frac{3L^2}{4}$ $\endgroup$ – Vasya Sep 4 '18 at 19:16
  • $\begingroup$ You can get properly sized parentheses that adjust to their content by preceding them with \left and \right. Please see this tutorial and reference on how to typeset math on this site. $\endgroup$ – joriki Sep 4 '18 at 20:08
  • $\begingroup$ @Cedric Martens I corrected a stupid mistake in my answer. (Tired when I posted, I forgot we had $2$ right triangles. Please look at it again and see if it helps. $\endgroup$ – poetasis May 22 at 16:33
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With the Pythagorean theorem, you can find the altitude of an equilateral triangle by dropping a vertical line to split it. Then long side and half of the bottom are, respectively, the hypotenuse $C$ and a short leg $B$ of a right triangle. The vertical will be side $A$ of a right triangle where $A=\sqrt{C^2-\bigl{(}\frac{C}{2}\bigr{)}^2}=\sqrt{\frac{4C^2-C^2}{2}}=\frac{C\sqrt{3}}{2}$. The area of one triangle is $\frac{1}{2}A*B$ where $B=\frac{C}{2}$. However, we have two of these right triangles so the area is simply A*B. You should end up with $area=A*2B=AC=\frac{C\sqrt{3}}{2}*C=\frac{C^2\sqrt{3}}{2}$.

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