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Calculating the area of equilateral triangle using calculus

  1. Jul 23, 2017 #1
    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
    equilateral T.png
    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. jcsd
  3. Jul 23, 2017 #2

    scottdave

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    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.
     
  4. Jul 23, 2017 #3
    It's so much easier to integrate the area of the triangle directly.
     
  5. Jul 23, 2017 #4

    scottdave

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