We can divide the snowflake into a big square and a 2 small squares as:
The big square has side c with measure
[tex]3^2+3^2=c^2[/tex]
which was obtained from the following right triangle:
Then, c (the lenght of the square sides) is given by
[tex]\begin{gathered} c=\sqrt[]{3^2+3^2} \\ c=\sqrt[]{9+9} \\ c=\sqrt[]{2\cdot9} \\ c=3\sqrt[]{2} \end{gathered}[/tex]
Therefore, the area of the big square is
[tex]\begin{gathered} A_{\text{Big}}=c^2 \\ A_{\text{Big}}=(3\sqrt[]{2})^2 \\ A_{\text{Big}}=9\cdot2 \\ A_{\text{Big}}=18units^2 \end{gathered}[/tex]
Now, the sides of the small squares measure 1 unit, then the area of one small square is
[tex]\begin{gathered} A_{\text{small}}=1^2 \\ A_{\text{small}}=1units^2 \end{gathered}[/tex]
Finally, the total area is
[tex]A=A_{\text{big}}+4\cdot A_{\text{small}}[/tex]
By substituting the last results, we get
[tex]\begin{gathered} A=18+4\cdot1 \\ A=18+4 \\ A=22units^2 \end{gathered}[/tex]
that is, the area of the snowflake is 22 units^2.
