I found this area

The area of the incircle of a right triangle is given by $\frac{π(h-b)(h-p)}{2}$ , where h, p and b are the hypotenuse, height and base of the right triangle respectively.

To solve this, I had to assume that what is referred to as "base" and "perpendicular" are the legs of the right triangle in question.

Relationships between the hypotenuse $h$ , base $b$ , and perpendicular $p$ are: $h-b=50, h-p=1$ , and $p^2+b^2=h^2.$

Solving for $h$ we get two solutions: $h=61$ and $h=41$ . Calculating $b$ from the second one of those will give us a negative number, so this is not geometrically reasonable.

Using the first solution, we have $b=11, p=60$ , the area of the triangle $A=\frac{1}{2}pb=330$ , and semiperimeter $s=\frac{1}{2}(b+p+h)=66$ .

Radius of incircle $R=\frac{A}{s}=5$ . Area of incircle is therefore $25\pi$ .