# Visible confusion - hypotenuse for the triangle is 4 sqrt 2?

• How does Prof. Loh in the beginning already know that the hypotenuse for the triangle is 4 sqrt 2?

• And how does he know that the radius of the entire semi-circle (the unshaded and shaded thing) is 2 sqrt 2?

• And how does that method at the end work? This is proving to be extremely confusing to me

• 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