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Difference between revisions of "Area of an equilateral triangle"

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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>\frac{1}{2}bh</math> (note <math>s</math> is the length of a base), so the area is <cmath>\frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) \boxed{\frac{s^2\sqrt{3}}{4}}</cmath>
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We use the formula for the area of a triangle, <math>\frac{1}{2}bh</math> (note that <math>s</math> is the length of a base), so the area is <cmath>\frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) = \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^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</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^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</math>

Latest revision as of 03:06, 18 June 2018

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, $\frac{1}{2}bh$ (note that $s$ is the length of a base), so the area is \[\frac{1}{2}(s)\left(\frac{s\sqrt{3}}{2}\right) = \boxed{\frac{s^2\sqrt{3}}{4}}\]

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 $\frac{s^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}$

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