Differentiate a triangle constrained to fixed area

$ab/2=100$ gives $b=200/a$, with $a>0$. Thus $c^2=a^2+200^2/a^2$. When $a=20$, $b=10$ and $c=\sqrt{500}=10\sqrt5$.

Differentiate the fixed-area relation after substitution:

$\begin{aligned}&2c\frac{dc}{dt}\\ & =\left(2a-\frac{2(200)^2}{a^3}\right)\frac{da}{dt}.\end{aligned}$

At $a=20$ and $da/dt=3$, the right-hand side is $(40-10)3=90$. Hence $dc/dt=90/(20\sqrt5)=9/(2\sqrt5)\text{ cm s}^{-1}\approx2.01\text{ cm s}^{-1}$, so the hypotenuse is increasing.

With $c=(a^2+200^2/a^2)^{1/2}$,

$\begin{aligned}&\frac{dc}{da}\\ & =\frac12\left(a^2+\frac{200^2}{a^2}\right)^{-1/2}\left(2a-\frac{2(200)^2}{a^3}\right).\end{aligned}$

Use $dc/dt=(dc/da)(da/dt)$ and substitute $a=20$ to obtain the same $9/(2\sqrt5)\text{ cm s}^{-1}$, or the printed $2.01\text{ cm s}^{-1}$.