How fast is the area increasing for an equilateral triangle under the given conditions? Question:. 
 The sides of an equilateral triangle are increasing at the rate of $\sqrt{3}$ cm/sec. How fast is the area increasing when its side is $6$ cm?   
I have done this problem using calculus.    Is there any alternative way to solve the above problem? 
Given the sides the equilateral triangle = $a$ cm.
 Area of the equilateral triangle is 
 $A=\dfrac{\sqrt{3}}{4}a^2$
 differentiating w.r.t $t$ we get,
$\dfrac{dA}{dt}=\dfrac{\sqrt{3}}{4}\times 2a\times \dfrac{da}{dt}$
 But it is given by
 $\dfrac{da}{dt}=\sqrt{3}$ cm/sec
 when $a=6$ cm.
$\therefore \left[\dfrac{dA}{dt}\right]_{a=6}=\dfrac{\sqrt{3}}{4}\times 2\times 6\times \sqrt{3}=9\; cm^2/sec$.

This still uses calculus as its underlying mechanic, but we can pretend we don't know any of it. We transform the problem into a physic problem.
An object is accelerating at constant $a = \sqrt{3}\ \text{cm/sec}^2$. Using the formula for constant acceleration we find that for initial velocity $0$ that $x = \frac{1}{2}at^2$ where $x$ is the distance traveled. We know $x = 6\ \text{cm}$ and $a = \sqrt{3}\ \text{cm/sec}^2$ so we solve for $t$:
$$6 = \frac{1}{2} \cdot \sqrt 3 \cdot t^2$$
$$t = \sqrt{\frac{2 \cdot 6}{\sqrt 3}} = 2\sqrt[4]{3}$$
Now we can calculate the velocity at $t = 2\sqrt[4]{3}$ using $v = at$ to find $v = 2 \cdot 3^{3/4}$.
So we know that the side of the equilateral triangle when it reaches $6\ \text{cm}$ is increasing by $2 \cdot 3^{3/4}\ \text{cm/sec}$, and thus the area of the equilateral triangle is increasing by $9\ \text{cm}^2/\text{sec}$
