Leg gets larger

One leg of the right triangle is of $1$ unit. Let the other leg be of $n$ units.Then its hypotenuse will be of $\sqrt{n^2 + 1}$ . Let the inradius of right triangle be $r$ .

We know that area of right triangle is half the product of its legs. Also the area of any triangle is inradius times semiperimeter. Applying these two concept :

Semiperemeter of right triangle $= \large\frac{n + 1 + \sqrt{n^2 + 1}}{2}$

$\frac{1}{2}\cdot (n)(1) = r\big( \large\frac{n + 1 + \sqrt{n^2 + 1}}{2}\big)$

$\Rightarrow r = \large\frac{n}{n + 1 + \sqrt{n^2 + 1}}$

$\Rightarrow r = \large\frac{1}{1 + \frac{1}{n} + \sqrt{1 + \frac{1}{n^2}}}$

$\lim_{n \to \infty} r = \lim_{n \to \infty} \large\frac{1}{1 + \frac{1}{n} + \sqrt{1 + \frac{1}{n^2}}}$ $= \large\frac{1}{2}$

Area of inscribed circle is $\pi r^2 = \pi(\frac{1}{2})^2 = \large\frac{\pi}{4}$ $\approx 0.785$

Let the length of the other leg be x units. Then, by the Pythagorean Theorem, the length of the hypotenuse is (1+x^2)^.5. The inscribed circle is tangent to each leg and the hypotenuse. Draw 3 radii from the center of the circle to each point of tangency. Partition the big triangle into three smaller triangles such that the height of each triangle is the radius of the circle. We then see, if R is the radius:

Area of the Big Triangle = (1/2) (BASE) (HEIGHT) = x/2

Area of the Big Triangle = Sum of the Areas of the Three Little Triangles = (1/2) R + (1/2) R x + (1/2) R*(1+x^2)^.5.

Setting these expressions equal and solving for R gives:

R = x (1+x+(x^2+1)^.5)^-1. Letting x approach infinity, R approaches 1/2. The radius of the circle, which is Pi R^2, then approaches Pi / 4, which is approximately .785.