Find the least area this pentagon can have.

We can use Pick's formula to make analysis a bit easier. If $i$ is the number of interior points, and $b$ is the number of vertices, then the area is

$i+\frac{1}{2}b-1$

The least this can be for $b=5$ would be $\frac{3}{2}$ , but only if such a pentagon can be drawn without any interior points. Such a pentagon can be drawn, but it will not be convex.

The next least is $\frac{5}{2}$ , where there is $1$ interior point, as you have drawn.

Take a look at the grid which shows a simple illustration, with all integer points.

Then the area of the required pentagon is 1 + 0.5 + 0.5 + 0.5 = 2.5