Given: $a^{2}, b^{2}$ and $1 c^{2}$ are squares of consecutive integers. If $a^{2}$ and $c^{2}$ are known, without directly calculating the square roots of $a^{2}$ or $c^{2}$, determine the value of $b^{2}$.

Without using the calculation of the square roots of $a^{2}$ or $c^{2}$, find the value of $b^{2}$.
[Method 1] Given: $b^{2}=\frac{a^{2}+c^{2}}{2}-1$,
think of $c=a+2$. Then,
$$
\begin{array}{l}
4^{2}+\frac{(a+2)^{2}}{2}-1=\frac{a^{2}+a^{2}+4 a+4}{2}-1 \\
=\frac{2 a^{2}+4 a+4}{2}-1=a^{2}+2 a+1=(a+1)^{2}
\end{array}
$$

By definition, this is $b^{2}$.
[Method 2] Prove: $b^{2}=\left(\frac{c^{2}-a^{2}}{4}\right)^{2}$, think of $c=a+2$, then,
$=\left(\frac{4 a}{4}+4\right)^{2}=(a+1)^{2}$. By definition, this is $b^{2}$.
[Method 3] Prove: $b^{2}=a^{2}$
$+\left(\frac{c^{2}-a^{2}}{2}\right)-1$, think of $c=a+2$, then,
$$
\begin{array}{l}
a^{2}+\left(\frac{(a+2)^{2}-a^{2}}{2}\right)-1=a^{2} \\
+\left(a^{2}+\frac{4 a+4-a^{2}}{2}\right)-1=a^{2}+2 a+2-1 \\
=a^{2}+2 a+1=(a+1)^{2} . \text { This is } b^{2} .
\end{array}
$$

Verify that all these transformations can be geometrically explained using the area of a rectangle.
For example, use the following diagram to verify transformation 1, which is
$$
b^{2}=\frac{a^{2}+c^{2}}{2}-1
$$