Hi again! Let me answer all of your questions.

For your first one, which set of lines are you referring to? The two yellow line segments are the same because they are both radii of the circle, and the two white segments are the same because, as Professor Loh said, you can apply the Pythagorean Theorem!

Similar triangles are very different from congruent triangles. Two triangles are congruent if they are exactly the same! (Though, it still counts if it's "flipped"). The important thing about congruent triangles is that they share the exact same side lengths and angles. This is very useful: If we can figure out that two triangles are congruent, then we may be able to deduce that two angles are the same, or perhaps that two side lengths are the same!

On the other hand, similar triangles don't have as much in common as congruent triangles. Similar triangles only need the same angles. A good way to think about what triangles count as similar is this: If you draw a triangle, and draw the "same" triangle but 3 times smaller, you will have two triangles that both "look" the same, but they aren't congruent because the second triangle has sides that are three times smaller. But, they will actually be similar!

Finally let's get to your last question, an algebra question! Why is it that:

$$\sqrt{2} \times \sqrt{4} = 2 \times \sqrt{2}?$$

Try not to overthink this! We know that \(\sqrt{4} = 2\), so \(\sqrt{2} \times \sqrt{4} = \sqrt{2} \times 2 = 2 \times \sqrt{2}\).

Though, in the future, dealing with radicals might be tricky! For example, how might you simplify this?

$$\sqrt{8} \times \sqrt{12}$$

Here's one approach: First we can combine the radicals together:

$$\sqrt{8} \times \sqrt{12} = \sqrt{8 \times 12}$$

Now, look for perfect squares you can separate it into, so you can take them out!

$$\sqrt{8 \times 12} = \sqrt{4 \times 2 \times 12} = \sqrt{4 \times 4 \times 6} = \sqrt{4} \times \sqrt{4} \times \sqrt{6} = \boxed{4\sqrt{6}}$$

Let me know if you have any other questions! (Unfortunately, I can't answer your question about course prerequisites, but I've asked the other team members about it, so you'll get an answer soon!)

Happy learning!