Difference between revisions of "Area of an equilateral triangle"
(→Proof) |
(→Proof) |
||
| Line 14: | Line 14: | ||
---- | ---- | ||
| − | credits: | + | |
| + | == credits: == | ||
created by [[User:Pickten|Pickten]] 18:22, 12 June 2011 (EDT) | created by [[User:Pickten|Pickten]] 18:22, 12 June 2011 (EDT) | ||
Revision as of 17:23, 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 ![\[\boxed{\frac{s^2\sqrt{3}}{4}}\]](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)
credits:
created by Pickten 18:22, 12 June 2011 (EDT)
