Answer
The area of the triangle is increasing at a rate of $~~150~\sqrt{3}~cm^2/min$
Work Step by Step
We can differentiate both sides of the equation of the triangle's area with respect to $t$:
$A = \frac{s^2~\sqrt{3}}{4}$
$\frac{dA}{dt} = \frac{s~\sqrt{3}}{2}~\frac{ds}{dt}$
$\frac{dA}{dt} = \frac{(30~cm)~\sqrt{3}}{2}~(10~cm/min)$
$\frac{dA}{dt} = 150~\sqrt{3}~cm^2/min$
The area of the triangle is increasing at a rate of $~~150~\sqrt{3}~cm^2/min$
