Answer
$(8x^{2}+32x+32)\ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}.$
Drawing a diagram helps in visualizing which area needs to be subtracted (which is outer, which is inner)
Area of red part=(Area of outer square) - (Area of inner square)
Work Step by Step
Find the total area of the inner square.
$(x+2)^{2}$
...square the binomial.
$=x^{2}+2(x)(2)+2^{2}$
...simplify.
$=\color{red}{x^{2}+4x+4}$
The length of the side of the outer square is $3\cdot(x+2)$
Find the area of the outer square.
$[3\cdot(x+2)]^{2}=3^{2}\cdot(x+2)^{2}$
...square the binomial.
$=9\cdot[x^{2}+2(x)(2)+2^{2}]$
...simplify.
$=9(x^{2}+4x+4)$
$=\color{red}{9x^{2}+36x+36}$
Find the area of the red part of the logo.
Area of figure=(Area of outer square) - (Area of inner square)
$A=9x^{2}+36x+36-(x^{2}+4x+4)$
...remove parentheses.
$=9x^{2}+36x+36-x^{2}-4x-4$
...group like terms.
$=9x^{2}-x^{2}+36x-4x+36-4$
...add like terms.
$=8x^{2}+32x+32$
The area of the red part of the logo is $(8x^{2}+32x+32)\ \mathrm{u}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{s}^{2}.$
