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

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== Proof == | == 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 <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.) | ||

− | We use the formula for the area of a triangle, <math>{ | + | 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\ | + | ''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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## Latest revision as of 03:06, 18 June 2018

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 that 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