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\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 17:29, 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 ![\[\boxed{\frac{s^2\sqrt{3}}{4}}\]](http://latex.artofproblemsolving.com/d/a/1/da1c83d5ea401ba837c9996377755e3bc0c0476a.png)
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. Unknown error_msg)
