Difference between revisions of "Area of an equilateral triangle"
m (Add some details) |
m |
||
Line 8: | Line 8: | ||
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>\frac{1}{2}bh</math> (note <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> | + | 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^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</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^2\sin{c}}{2}=\frac{s^2\frac{\sqrt{3}}{2}}{2}=\boxed{\frac{s^2\sqrt{3}}{4}}</math> |
Latest revision as of 04:06, 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