Answer
$(6x+9)\ \ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}$.
Work Step by Step
Find the total area of the outer square.
$(x+3)^{2}$
...square the binomial.
$=x^{2}+2(x)(3)+3^{2}$
...simplify.
$=x^{2}+6x+9$
Find the area of the inner square.
$x\cdot x=x^{2}$
Find the area of the figure
Area of figure=(Area of outer square) - (Area of inner square)
$A=x^{2}+6x+9-x^{2}$
...group like terms.
$=x^{2}-x^{2}+6x+9$
...add like terms.
$=6x+9$
The area of the figure is $(6x+9)\ \ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}$.
