Another way to see that a lune is nice.

We can let the side of the isosceles triangle be \(x\), and then we can use the same strategy as the first part explained. We can find the lune by doing
$$\text{area of the semicircle}(\text{area of the quarter circle}\text{area of the isosceles triangle}). $$Let's start with the area of the isosceles triangle. That is just \(\dfrac{x^2}{2}.\) Now we need to find the area of the quarter circle. Notice that the quarter circle has radius \(x\). By using the \(\pi r^2\) formula, we get \(\dfrac{x^2}{4}\pi\). In order to find the area of the semicircle, we need to find the length of the hypotenuse. We can scale the triangle with side lengths 1,1,\(\sqrt{2}\) by \(x\) to find that the hypotenuse is \(x\sqrt{2}\). From this, we know that the area of the semicircle is \(\dfrac{(x\sqrt{2})^2}{2}\pi=x^2\pi\). But after subtracting this doesn't imply that the answer is always nice. What did I do wrong?

@professionalbronco You're really close! The only mistake is the area of the semicircle; you're correct that the hypotenuse is x * sqrt(2). However, if you look close, the hypotenuse is actually the diameter of the semicircle... the radius would be half of that, or x * sqrt(2)/2. Then, the area of the semicircle comes out to be (x^2)pi/4, and the area of the lune is pi*(x^2)/4  pi * (x^2/4) + x^2 / 2 = x^2 / 2, exactly the area of the isosceles triangle!

@audrey thanks for pointing out my mistake! I didn't see that!

@professionalbronco No problem! Great job generalizing the result to all isosceles right triangles