Since what we are presented with is an isosceles triangle, the missing side will be equal to the side provided by the exercise. Also, remember that the area of a rectangle is given by
[tex]A_T=\frac{b\times h}{2}[/tex]
Where,
-b: the base of the triangle.
-h: the height of the triangle.
In that sense, the area of the triangle given is
[tex]\begin{gathered} A_T=\frac{(3g^2-6g+2)(3^g^2-6g+2)}{2} \\ A_T=\frac{1}{2}{}\lbrack9g^4-18g^3+6g^2-18g^3+36g^2-12g+6g^2-12g+4\rbrack \\ A_T=\frac{1}{2}(9g^4-36g^3+48g^2-24g+4) \\ A_T=\frac{9}{2}g^4-18g^3+24g^2-12g+2 \end{gathered}[/tex]
The standard form polynomial expression is given by the last equation in the above process. In other words, the answer is
9/2g^4-18g^3+24g^2-12g+2
