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Difference between revisions of "Area of an equilateral triangle"

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== Proof ==
 
== Proof ==
  
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''Method 1'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is <math>\frac{s}{2}</math>.
+
''Method 1:'' Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is <math>\frac{s}{2}</math>.
  
 
Using the [[Pythagorean theorem]], we get <math>s^2=h^2+\frac{s^2}{4}</math>, where <math>h</math> is the height of the triangle. Solving, <math>h=\frac{s\sqrt{3}}{2}</math>. (note we could use 30-60-90 right triangles.)
 
Using the [[Pythagorean theorem]], we get <math>s^2=h^2+\frac{s^2}{4}</math>, where <math>h</math> is the height of the triangle. Solving, <math>h=\frac{s\sqrt{3}}{2}</math>. (note we could use 30-60-90 right triangles.)
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We use the formula for the area of a triangle, <math>{bh \over 2}</math> (note <math>s</math> is the length of a base), so the area is <cmath>\boxed{\frac{s^2\sqrt{3}}{4}}</cmath>
 
We use the formula for the area of a triangle, <math>{bh \over 2}</math> (note <math>s</math> is the length of a base), so the area is <cmath>\boxed{\frac{s^2\sqrt{3}}{4}}</cmath>
  
−
''Method 2'' '''(warning: uses trig.)''' The area of a triangle is <math>\frac{ab\sin{C}}{2}</math>. Plugging in <math>a=b=s</math> and <math>C=\frac{\pi}{3}</math> (the angle at each vertex, in radians), we get the area to be <math>\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=</math>
+
''Method 2:'' '''(warning: uses trig.)''' The area of a triangle is <math>\frac{ab\sin{C}}{2}</math>. Plugging in <math>a=b=s</math> and <math>C=\frac{\pi}{3}</math> (the angle at each vertex, in radians), we get the area to be <math>\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=</math>
 
<cmath>\boxed{\frac{s^2\sqrt{3}}{4}}</cmath>
 
<cmath>\boxed{\frac{s^2\sqrt{3}}{4}}</cmath>

Revision as of 16:30, 10 November 2017

The area of an equilateral triangle is $\frac{s^2\sqrt{3}}{4}$, where $s$ is the sidelength of the triangle.


Proof

Method 1: Dropping the altitude of our triangle splits it into two triangles. By HL congruence, these are congruent, so the "short side" is $\frac{s}{2}$.

Using the Pythagorean theorem, we get $s^2=h^2+\frac{s^2}{4}$, where $h$ is the height of the triangle. Solving, $h=\frac{s\sqrt{3}}{2}$. (note we could use 30-60-90 right triangles.)

We use the formula for the area of a triangle, ${bh \over 2}$ (note $s$ is the length of a base), so the area is \[\boxed{\frac{s^2\sqrt{3}}{4}}\]

Method 2: (warning: uses trig.) The area of a triangle is $\frac{ab\sin{C}}{2}$. Plugging in $a=b=s$ and $C=\frac{\pi}{3}$ (the angle at each vertex, in radians), we get the area to be $\frac{s\cdot s\cdot \frac{\sqrt{3}{2}}{2}=$ (Error compiling LaTeX. Unknown error_msg) \[\boxed{\frac{s^2\sqrt{3}}{4}}\]

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