Rate of change of the area of a triangle

Let $|BC| = x.$ Then $|AC| = 2*|AB| - x = 4 - x.$ Thus, using Heron's formula , we see that

$S = \sqrt{3(3 - 2)(3 - x)(3 - (4 - x))} =$

$\sqrt{3(3 - x)(x - 1)} = \sqrt{3}\sqrt{-x^{2} + 4x - 3},$ (i).

Now we also have that $S = \dfrac{1}{2}|AB||BC|\sin(\theta) = x\sin(\theta),$ and so

$\sin(\theta) = \dfrac{S}{x} = \dfrac{\sqrt{3}\sqrt{-x^{2} + 4x - 3}}{x}.$

Differentiating both sides with respect to $t$ gives us that

$\cos(\theta)\dfrac{d\theta}{dt} = \dfrac{\sqrt{3}(3 - 2x)}{x^{2}\sqrt{-x^{2} + 4x - 3}}\dfrac{dx}{dt},$

and since $\dfrac{d\theta}{dt} = 2$ rad\sec we find that

$\dfrac{dx}{dt} = \dfrac{2x^{2}\cos(\theta)\sqrt{-x^{2} + 4x - 3}}{\sqrt{3}(3 - 2x)}.$

Differentiating equation (i) with respect to $t$ using the chain rule, we have that

$\dfrac{dS}{dt} = \dfrac{dS}{dx}*\dfrac{dx}{dt} =\dfrac{\sqrt{3}(2 - x)}{\sqrt{-x^{2} + 4x - 3}}*\dfrac{2x^{2}\cos(\theta)\sqrt{-x^{2} + 4x - 3}}{\sqrt{3}(3 - 2x)} =$

$\dfrac{2x^{2}(2 - x)\cos(\theta)}{3 - 2x},$ (ii).

Now when $\angle A = \dfrac{\pi}{2}$ we have that, by Pythagoras,

$|BC|^{2} = |AB|^{2} + |CA|^{2} \Longrightarrow x^{2} = 4 + (4 - x)^{2} \Longrightarrow 8x = 20 \Longrightarrow x = \dfrac{5}{2}.$

This then gives us that $\cos(\theta) = \dfrac{|AB|}{|BC|} = \dfrac{4}{5}.$

Plugging these values into equation (ii) gives us that, when $\angle A = \dfrac{\pi}{2},$

$\dfrac{dS}{dt} = \dfrac{2*\dfrac{25}{4}*(-\dfrac{1}{2})*\dfrac{4}{5}}{3 - 5} = \dfrac{5}{2} = \boxed{2.5}$ units/sec.