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

**sluggerbroth** · May 2nd 2012, 06:04 AM

**Prove It** · May 2nd 2012, 06:17 AM

Originally Posted by sluggerbroth

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 $\displaystyle \begin{align*} a = b = c = s \end{align*}$ .

**biffboy** · May 2nd 2012, 06:49 AM

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.

**Soroban** · May 2nd 2012, 06:50 AM

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  :

$\text{The base of the triangle is }s.$
$\text{The height is }h = AD.$

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

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

$\text{Therefore: }\:A \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}(s)\left(\tfrac{\sqrt{3}}{2}s \right) \;=\;\frac{\sqrt{3}}{4}s^2$

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