@the-blade-dancer Thank you 🙂 We will fix this someday 🙂

Hello,

To help illustrate why the hypotenuse is 4(sqrt 2), think about a right triangle with legs 1 and 1. By the Pythagorean Theorem, we know the hypotenuse is (sqrt 2). So if we scale this triangle by 4, as it is in the problem, the hypotenuse should have length 4(sqrt 2).

To answer your other question about the radius of the semi-circle, we can see that the whole hypotenuse is the diameter of the semi-circle, as shown in the picture. Given that the radius is half of the diameter, and the diameter is 4(sqrt 2), then the radius is just 2 (sqrt 2)!

I hope that helps. Let us know if you have any other questions!

Happy Learning,

The Daily Challenge Team

Adding the word "respectively" would, without a doubt, clarify which center belongs to which circle. However, you can also use some intuition to figure this out. It would take more time, but the fun of proving something is worth it!

Suppose we call the two arcs "inner" and "outer." Supposing the "inner" arc has its center as the midpoint of BC. Then BC would be a diameter of this circle. From inspection, this looks weird as it is, but you can also think about the fact that any radius should be perpendicular to a tangent line passing through its intersection point with the circle. In other words, if BC were the diameter of the "inner" arc, then it should be perpendicular to the tangent to the circle at point C. (This tangent line, by the way, appears horizontal, parallel to AB.) This tangent line, though is perpendicular to a different line, AC. This also implies that a radius to the circle containing the "inner" arc should pass through AC! And that happens to bring us to point A. So this tell us that A must be the center of the "inner" arc.

Please let us know if this was still unclear, or if you have any more questions. It's my pleasure to help, anytime!

Happy Learning,

The Daily Challenge Team