## Intermediate Algebra (12th Edition)

$A=\dfrac{9x^2-4y^2}{2}\text{ square units}$
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}