Answer
$A=\dfrac{9x^2-4y^2}{2}\text{ square units}$
Work Step by Step
Using $A=\dfrac{1}{2}bh$, the area of the triangle with height equal to $
3x-2y
$ and base equal to $
3x+2y
,$ is
\begin{array}{l}\require{cancel}
A=\dfrac{1}{2}(3x-2y)(3x+2y)
.\end{array}
Using $(a+b)(a-b)=a^2-b^2$ or the product of the sum and difference of like terms, the expression above is equivalent to
\begin{array}{l}\require{cancel}
A=\dfrac{1}{2}[(3x)^2-(2y)^2]
\\\\
A=\dfrac{1}{2}(9x^2-4y^2)
\\\\
A=\dfrac{9x^2-4y^2}{2}\text{ square units}
.\end{array}