Find rate of change of the area of a triangle at a particular moment The side lengths of a right-angled triangle are $a, b$ and $c$ centimeters where $c$ is the length of the hypotenuse and $a, b$ and $c$ are changing as differentiable functions of time. It is known that the perimeter $a+b+c$ is always constant and at the moment $a=3$ and $b=4$ the rate of change of $a$ is $108$ cm/sec. What is the rate of change of the area of the triangle (in cm$^2$/sec units) at this moment?
This is a related rates question (the answer is $72$). What I tried and first of all I calculated the derivative of the area of a right triangle ($ab/2$) and everything is known except the derivative of side $b$. I need to find some relation between $b$ and $a$ but I am not sure how.

$\frac{d a}{dt} + \frac{d b}{dt} + \frac{d c}{dt} = 0$ ...(i)
$c^2 = a^2 + b^2$ and that gives you
$c \frac{d c}{dt} = a \frac{d a}{dt} + b \frac{d b}{dt}$
Now when $a = 3, b = 4$, $c = 5$
$5 \frac{d c}{dt} = 3 \frac{d a}{dt} + 4 \frac{d b}{dt}$ ...(ii)
From (i) and (ii), you can find $\frac{d b}{dt} = -96$ as you know $\frac{d a}{dt} = 108$.
Now $\frac{d A}{dt} = \frac{1}{2} \big(a \frac{d b}{dt} + b \frac{d a}{dt} \big)$
