Solution:

Let the area of the $n$th square drawn be $S_{n}$ and the area of the $n$th triangle be $T_{n}$. Since the hypotenuse of the $n$th triangle is of length $\sqrt{S_{n}}$, its legs are of length $l=\sqrt{\frac{S_{n}}{2}}$, so $S_{n+1}=l^{2}=\frac{S_{n}}{2}$ and $T_{n}=\frac{l^{2}}{2}=\frac{S_{n}}{4}$. Using the recursion relations, $S_{n}=\frac{1}{2^{n-1}}$ and $T_{n}=\frac{1}{2^{n+1}}$, so $S_{n}+T_{n}=\frac{1}{2^{n-1}}+\frac{1}{2^{n+1}}=\left(\frac{1}{2}+2\right) \frac{1}{2^{n}}=\frac{5}{2} \frac{1}{2^{n}}$. Thus the total area of the figure is $\sum_{n=1}^{\infty} S_{n}+T_{n}=\frac{5}{2} \sum_{n=1}^{\infty} \frac{1}{2^{n}}=\frac{5}{2}$.