express area of equilateral triangle as function of one side s???
Use Heron's formula $\displaystyle \displaystyle \begin{align*} A = \sqrt{S(S - a)(S - b)(S-c)} \end{align*}$ where S is the semiperimeter, so $\displaystyle \displaystyle \begin{align*} S = \frac{a + b + c}{2} \end{align*}$, and $\displaystyle \displaystyle \begin{align*} a = b = c = s \end{align*}$.
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.
Hello, sluggerbroth!
$\displaystyle \text{Express area of equilateral triangle as function of one side }s.$
$\displaystyle \text{The base of the triangle is }s.$Code:A * /|\ / | \ / | \ s / | \ s / |h \ / | \ / | \ B * - - - * - - - * C : s/2 D s/2 :
$\displaystyle \text{The height is }h = AD.$
$\displaystyle \text{In right triangle }ADC:\:h^2 + (\tfrac{s}{2})^2 \:=\:s^2 \quad\Rightarrow\quad h^2 + \tfrac{s^2}{4} \:=\:s^2$
. . $\displaystyle h^2 \:=\:\tfrac{3}{4}s^2 \quad\Rightarrow\quad h \:=\:\tfrac{\sqrt{3}}{2}s$
$\displaystyle \text{Therefore: }\:A \;=\;\tfrac{1}{2}bh \;=\;\tfrac{1}{2}(s)\left(\tfrac{\sqrt{3}}{2}s \right) \;=\;\frac{\sqrt{3}}{4}s^2$