Pythagoras, inscribed

The sides are Pythagorean triples. It always have two odd and one leg even or all even sides. So its area, half product of the legs, is an integer. Its semi-perimeter too is half of two odd and one even side or all even sides, so it is an integer. With usual notations, and noting that $a^2+b^2=c^2$ . {Thanks to Deeparaj Bhat's solution for part of below solution marked blue.}
$A_i=\dfrac {\frac 1 2 \cdot a \cdot b}{ \frac 1 2 (a+b+c) }=\color{#3D99F6}{\dfrac {(a \cdot b)(a+b - c)}{ (a+b+c)(a+b - c)} }=\dfrac {(a \cdot b)(a+b - c)}{ (a+b)^2 - c^2}\\ =\dfrac {(a \cdot b)(a+b - c)}{ 2ab}=\dfrac {a+b - c} 2.$ $a+b -c$ is difference of two odd plus an even, or all even, is always an even. $\implies\ A_i=\dfrac {a+b - c} 2$ is an integer. For triple 6-8-10, $A_i=2$ , not a perfect square.

The angle at C is a right angle, and let the radius of the inscribed circle be r. The length of the tangent from B to the circle is a - r, and from A is b - r. Therefore c = a - r + b - r and hence 2r = a + b - c. The sides are of integer length and so we have a Pythagorian triple (2mn, m^2 - n^2, m^2 + n^2) and r = n(m - n) which is an integer. The triple (5, 12, 13), with m = 3 and n = 2, shows r is not always a square.

Note that $A_i=\dfrac{\Delta}{s}$ where $\Delta$ denotes the area of the triangle and $s$ is the semi-perimeter of the triangle (see incircle of triangle ) . Then, upon a little simplification, $\begin{aligned} A_i&=\frac{AC\cdot BC}{AB+BC+CA}\\ &=\frac{(AC\cdot BC)(AC+BC-AB)}{(AC+BC)^2-AB^2}\\&=\frac{AC+BC-AB}{2}\end{aligned}$ Note that either all the side lengths are even or only one side length is even, hence $A_i$ is an integer.

To show that it need not be a perfect square, take $AB=25, \, AC=24, \, BC=7$