# Difference between revisions of "Area of an equilateral triangle"

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''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> | ||

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## Revision as of 21:54, 12 June 2011

The area of an equilateral triangle is , where 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 .

Using the Pythagorean theorem, we get , where is the height of the triangle. Solving, . (note we could use 30-60-90 right triangles.)

We use the formula for the area of a triangle, (note is the length of a base), so the area is

*Method 2* **warning: uses trig.** The area of a triangle is . Plugging in and (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. ! File ended while scanning use of \frac .)