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

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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\ | + | ''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> |

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## Revision as of 17:40, 10 November 2017

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