+91 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.
+1223 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.
\(\frac{2 \cdot 2}{2} + 2 + \frac{3 \cdot 1}{2} + 1 + \frac{2 \cdot 1}{2} + \frac{3 \cdot 1}{2}\)
= \(2 + 2 + \frac{3}{2} + 1 + 1 + \frac{3}{2}\)
= \(\boxed{9}\)
