Results 1 to 4 of 4

Math Help - express area of equilateral triangle as function of one side s???

  1. #1
    Member sluggerbroth's Avatar
    Joined
    Apr 2012
    From
    Chicago
    Posts
    108

    express area of equilateral triangle as function of one side s???

    express area of equilateral triangle as function of one side s???
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,653
    Thanks
    1478

    Re: express area of equilateral triangle as function of one side s???

    Quote Originally Posted by sluggerbroth View Post
    express area of equilateral triangle as function of one side s???
    Use Heron's formula \displaystyle \begin{align*} A = \sqrt{S(S - a)(S - b)(S-c)} \end{align*} where S is the semiperimeter, so \displaystyle \begin{align*} S = \frac{a + b + c}{2} \end{align*}, and \displaystyle \begin{align*} a = b = c = s \end{align*}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Mar 2012
    From
    Sheffield England
    Posts
    440
    Thanks
    76

    Re: express area of equilateral triangle as function of one side s???

    Let h = height of triangle. Then h=s*sin60 So area= 1/2*s*s*sin60=1/2*s^2*sin60 Sin60=root3/2 so area = root3*s^2/4
    Can get the height by 'splitting' triangle in two and using Pythagoras, if you prefer.
    Last edited by biffboy; May 2nd 2012 at 06:53 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,803
    Thanks
    692

    Re: express area of equilateral triangle as function of one side s???

    Hello, sluggerbroth!

    \text{Express area of equilateral triangle as function of one side }s.

    Code:
                  A
                  *
                 /|\
                / | \
               /  |  \
            s /   |   \ s
             /    |h   \
            /     |     \
           /      |      \
        B * - - - * - - - * C
          :  s/2  D  s/2  :
    \text{The base of the triangle is }s.
    \text{The height is }h = AD.

    \text{In right triangle }ADC:\:h^2 + (\tfrac{s}{2})^2 \:=\:s^2 \quad\Rightarrow\quad h^2 + \tfrac{s^2}{4} \:=\:s^2

    . . h^2 \:=\:\tfrac{3}{4}s^2 \quad\Rightarrow\quad h \:=\:\tfrac{\sqrt{3}}{2}s


    \text{Therefore: }\:A \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}(s)\left(\tfrac{\sqrt{3}}{2}s \right) \;=\;\frac{\sqrt{3}}{4}s^2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Area - circles and equilateral triangle.
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 14th 2010, 03:38 AM
  2. area of an equilateral triangle
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 31st 2010, 10:02 AM
  3. Replies: 3
    Last Post: October 29th 2009, 07:53 AM
  4. Replies: 3
    Last Post: September 5th 2009, 02:30 PM
  5. Replies: 1
    Last Post: October 28th 2008, 07:02 PM

Search Tags


/mathhelpforum @mathhelpforum