A square wire of side $3.0 \mathrm{cm}$ is placed $25 cm$ from a concave mirror of focal length $10 cm$. The area enclosed by the image of the wire is

$u = - 25 cm$ and $f = - 10 cm$. The distance of the image is given by
$$\begin{gathered} \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\ =\frac{1}{-10}-\frac{1}{-25} \end{gathered}$$
or $v=-\frac{50}{3}cm$

Area of the object wire is ${\mathrm{A}}_{\mathrm{O}}=3.0 \times 3.0 = 9.0 {\mathrm{cm}}^2$. 

The area magnification is given by

$\mathrm{areal} \mathrm{magnification}=\frac{\mathrm{area} \mathrm{of} \mathrm{image}}{\mathrm{area} \mathrm{of} \mathrm{object}}={(\mathrm{linear} \mathrm{magnification})}^2$

$\frac{{\mathrm{A}}_{\mathrm{I}}}{{\mathrm{A}}_{\mathrm{O}}}=\frac{{\mathrm{v}}^2}{{\mathrm{u}}^2}={\left(\frac{50}{3\times 25}\right)}^2=\frac{4}{9}$

Therefore, the area enclosed by image is
${\mathrm{A}}_{\mathrm{I}}=\frac{4}{9}\times 9 {\mathrm{cm}}^2=4 {\mathrm{cm}}^2$