## Calculus 8th Edition

$A=\frac{\sqrt{3}}{4}x^2$, $x>0$
We have an equilateral triangle with sides $x$ and we need to find the area. We know that: $A=1/2*base*height=\frac{1}{2}xh$ We need to eliminate $h$. We construct a right triangle in the middle of the equilateral triangle with sides $h$, $x$, and $1/2x$. We use the Pythagorean Theorem with these three sides: $(\frac{1}{2}x)^2+h^2=x^2$ $h^2=x^2-(\frac{1}{2}x)^2$ $h=\pm\sqrt{x^2-(\frac{1}{2}x)^2}$ $h=+\sqrt{\frac{3}{4}x^2}=\frac{\sqrt{3}}{2}x$ (We eliminate the negative because lengths must be positive.) We plug in $h$ in the area formula: $A=\frac{1}{2}x\frac{\sqrt{3}}{2}x=\frac{\sqrt{3}}{4}x^2$ The domain is $x>0$ because lengths must be positive.