Prove AC^2 = 4 * sqrt(3) / 3 for Equilateral Triangle ABC

  • Thread starter Thread starter math_phys_bio
  • Start date Start date
  • Tags Tags
    Geometry
AI Thread Summary
To prove that AC^2 = 4 * sqrt(3) / 3 for an equilateral triangle ABC with an area of 1 square cm, start by noting that all sides are equal, denoted as x. The semi-perimeter s is calculated as s = (3x)/2. The area can be expressed using the formula area = {s(s-a)(s-b)(s-c)}^2, which simplifies to find x. Additionally, the height CC' can be expressed in terms of AC, allowing the use of the triangle area equation to derive the necessary relationship. The discussion also touches on the ratio theorem related to similar triangles, emphasizing the geometric properties involved.
math_phys_bio
Messages
3
Reaction score
0
ABC is an equilateral triangle with an area of 1 square cm.
C' is the middle of [AB].

i have to prove that AC^2 = 4 * sqrt(3) / 3

how?
 
Physics news on Phys.org
use s = (a+b+c)/2 and area = {s*(s-a)*(s-b)*(s-c)}^2
where a = b = c = x (say)
 
i know this aint helping but since the question is also on triangle can i ask how u prove ratio theorm?
 
ratio theorem ?? as in simillar triangles ??
 
Write CC' (triangle height) in terms of length AC, then solve using triangle area equation.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

Similar threads

Back
Top