To determine the area [tex]\( A \)[/tex] of polygons drawn with 3 to 10 boundary dots [tex]\( (b) \)[/tex] where there are only 2 interior dots [tex]\( (i) \)[/tex], we can use Pick's Theorem. Pick's Theorem states that for a simple polygon with [tex]\( i \)[/tex] interior lattice points and [tex]\( b \)[/tex] boundary lattice points, the area [tex]\( A \)[/tex] of the polygon can be given by:
[tex]\[
A = i + \frac{b}{2} - 1
\][/tex]
Given that [tex]\( i = 2 \)[/tex], we substitute into the formula as follows:
[tex]\[
A = 2 + \frac{b}{2} - 1 = \frac{b}{2} + 1
\][/tex]
Now, let's calculate [tex]\( A \)[/tex] for each value of [tex]\( b \)[/tex] from 3 to 10.
1. For [tex]\( b = 3 \)[/tex]:
[tex]\[
A = \frac{3}{2} + 1 = 1.5 + 1 = 2.5
\][/tex]
2. For [tex]\( b = 4 \)[/tex]:
[tex]\[
A = \frac{4}{2} + 1 = 2 + 1 = 3.0
\][/tex]
3. For [tex]\( b = 5 \)[/tex]:
[tex]\[
A = \frac{5}{2} + 1 = 2.5 + 1 = 3.5
\][/tex]
4. For [tex]\( b = 6 \)[/tex]:
[tex]\[
A = \frac{6}{2} + 1 = 3 + 1 = 4.0
\][/tex]
5. For [tex]\( b = 7 \)[/tex]:
[tex]\[
A = \frac{7}{2} + 1 = 3.5 + 1 = 4.5
\][/tex]
6. For [tex]\( b = 8 \)[/tex]:
[tex]\[
A = \frac{8}{2} + 1 = 4 + 1 = 5.0
\][/tex]
7. For [tex]\( b = 9 \)[/tex]:
[tex]\[
A = \frac{9}{2} + 1 = 4.5 + 1 = 5.5
\][/tex]
8. For [tex]\( b = 10 \)[/tex]:
[tex]\[
A = \frac{10}{2} + 1 = 5 + 1 = 6.0
\][/tex]
Fill in the table with the calculated areas:
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
(b) & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
(A) & 2.5 & 3.0 & 3.5 & 4.0 & 4.5 & 5.0 & 5.5 & 6.0 \\
\hline
\end{tabular}
So, the relationship between [tex]\( b \)[/tex] and [tex]\( A \)[/tex] when [tex]\( i = 2 \)[/tex] can be summarized by the formula:
[tex]\[
A = \frac{b}{2} + 1
\][/tex]
This gives us a straightforward method to determine the area for any polygon with [tex]\( i = 2 \)[/tex] interior dots by knowing only the boundary dots [tex]\( b \)[/tex].
