# Wed. Feb 22nd

Hi All —

In class today we solved a problem involving the area of triangles, circles and radicals. We took notes and practiced simplifying radicals, and students had time to work on the handout I gave yesterday on 8.4 and 8.5.

The area problem we worked on was to find the area of the shaded region.

As an example of where students are working to get by the end of the chapter, I had them solve this problem with me in notes.

To start breaking down this problem into some skills we can work on in class students need to be using equations in their process, simplifying radicals and drawing on background knowledge of properties and relationships in triangles.

We practiced simplifying radicals like $\dpi{300}\inline \sqrt{20}$

First, we find the prime factorization of 20: $\dpi{300}\inline 2 \cdot 2 \cdot 5$, so we can say

$\dpi{300}\inline \sqrt{20} = \sqrt{2} \cdot \sqrt{2} \cdot \sqrt{5}$

And since $\dpi{300}\inline \sqrt{2}$ is defined as the number which when multiplied by itself gives 2, $\dpi{300}\inline \sqrt{2} \cdot \sqrt{2} = 2$ So,,

$\dpi{300}\inline \sqrt{20} = 2 \cdot \sqrt{5}$

$\dpi{300}\inline 98 = 2 \cdot 49 = 2 \cdot 7 \cdot 7$, so $\dpi{300}\inline \sqrt{98} = 7\sqrt{2}$
$\dpi{300}\inline 600 = 6 \cdot 100$ so $\dpi{300}\inline \sqrt{600} = \sqrt{100} \cdot \sqrt{6} = 10\sqrt{6}$