# Calculating the area of equilateral triangle using calculus

1. Jul 23, 2017

1. The problem statement, all variables and given/known data
Calculating the area of equilateral triangle using calculus.

2. Relevant equations

3. The attempt at a solution

The area of the triangle is the area of the circle minus 3 times the area of the sector shown in (light blue). So, the target is to calculate the pink area first.
$y=\sqrt{r^2-x^2}$
The pink area is $\int_{0}^{x_0} \sqrt{r^2-x^2} dx$ = $\int_{0}^{x_0} r\sqrt{1-\frac{x^2}{r^2}} dx$ Putting $x=r sin a$ and doing the usual math with integration from 0 to $\pi/6$ led me to;
$r^2\int_{0}^{\pi/6} cos^2 a da$=$\frac{r^2}{2}\int_{0}^{\pi/6} cos (2a+1) da$=$\frac{r^2}{2} \left[\frac{sin2a}{2}+a\right]_0^{\pi/6}$=$\frac{r^2}{2} [\frac{1}{2}\frac{\sqrt 3}{2}+\pi/6]$=$\frac{r^2 \sqrt 3}{8} +\frac{\pi}{12}$
The blue area is then $2 (\frac{\pi r^2}{4}-\frac{r^2 \sqrt 3}{8} -\frac{\pi}{12})$=$\frac{\pi r^2}{2}-\frac{r^2 \sqrt 3}{4} -\frac{\pi}{6})$ and then the area of triangle is $\pi r^2-3(\frac{\pi r^2}{2}-\frac{r^2 \sqrt 3}{4} -\frac{\pi}{6})$
This will not give the correct result of $\frac{3\sqrt3 r^2}{4}$

2. Jul 23, 2017

### scottdave

Since you know the angle a = pi/6, you should be able to calculate the x value where the blue area starts, then just integrate sqrt(r2 - x2) from there to r. That gets you the top half of the blue area.

3. Jul 23, 2017

### willem2

It's so much easier to integrate the area of the triangle directly.

4. Jul 23, 2017

### scottdave

That too. With what he has figured, it should be easy to figure the equation of the line for the triangle.