Another way to see that a lune is nice.
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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?
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@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!
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@audrey thanks for pointing out my mistake! I didn't see that!
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@professionalbronco No problem! Great job generalizing the result to all isosceles right triangles
