#1**+2 **

We can use Pick's rule for the area:

\(A=I+{\frac {p} {2}}-1\)

-Where I is the points inside the figure and p is the points on the perimeter.

There are 5 points inside and 10 on the perimeter.

A = 5 + 10/2 - 1

A = 5+5-1

A=9

Therefore the area is **9 square units**.

ArithmeticBrains1234 Mar 19, 2021

#2**+2 **

If you don't know Pick's Theorem, you could also do it this way:

shaded = (area of total region) - (area not shaded)

= \((4 \cdot 4) - (\frac{2 \cdot 3}{2} + \frac{1 \cdot 1}{2} + 1 + \frac{1 \cdot 2}{2} + \frac{1 \cdot 3}{2})\)

= \(16 - 7\)

= \(\boxed{9}\)

You could also find the area of the shaded region directly, which is just made out of squares and triangles.

\(\frac{2 \cdot 2}{2} + 2 + \frac{3 \cdot 1}{2} + 1 + \frac{2 \cdot 1}{2} + \frac{3 \cdot 1}{2}\)

= \(2 + 2 + \frac{3}{2} + 1 + 1 + \frac{3}{2}\)

= \(\boxed{9}\)

CubeyThePenguin Mar 20, 2021