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Suppose the side length is $s$ cm, then the area is $s^2$ cm$^2$.

The radius of the circle is 1 cm, so the distance from the centre of the circle ($O$) to the vertices of the square that touch the circumference ($Q$ and $R$) is 1 cm.

This, the side length $s$ and the half-side length $\frac{1}{2}s$ are shown on the diagram:

Angle $OSR$ is a right angle (as it is in a square), so applying Pythagoras' theorem to triangle $OSR$ gives

$$\begin{align} s^2+(\frac{1}{2}s)^2 &= 1^2 \\

\Rightarrow s^2+\frac{1}{4}s^2 &=  1\\

\Rightarrow \frac{5}{4}s^2 &= 1\\

\Rightarrow s^2&=\frac{4}{5}\end{align}$$So the area of the square is $\dfrac{4}{5}$cm$^2$.