jmc

First, we shall sketch! The first step is to find $x.$ To do this, we simply plug into the Pythagorean Theorem: $$\begin{array}{rl}A{B}^2+B{C}^2& =A{C}^2\\ 90^2+x^2& =(2x-6{)}^2\\ 8100+x^2& =4x^2-24x+36\\ 0& =3x^2-24x-8064\\ 0& =x^2-8x-2688\\ 0& =(x-56)(x+48).\end{array}$$ The factorization is a little tricky, especially with a large constant term like $-2688,$ but it helps noticing that $2688$ is close to $52^2=2704,$ and the $-8x$ term indicates that our factors that multiply to $-2688$ have to be close. That helps narrow our search greatly.

In any case, clearly $x=-48$ is extraneous, so we have that $x=56.$ Therefore, we have $AC=106$ and $BC=56.$ (Did you know that $28:45:53$ is a Pythagorean triple?)

Now, to find the area of $△ADC$ is straightforward. First, clearly the height to base $DC$ is $90,$ so we only really need to find $DC.$ Here we use the Angle Bisector Theorem: $$\begin{array}{rl}\frac{BD}{DC}& =\frac{AB}{AC}\\ \frac{BD}{DC}& =\frac{90}{106}=\frac{45}{53}\\ 1+\frac{BD}{DC}& =1+\frac{45}{53}\\ \frac{BD+DC}{DC}=\frac{BC}{DC}& =\frac{98}{53}\\ \frac{56}{DC}& =\frac{98}{53}\\ DC& =\frac{53}{98}\cdot 56=\frac{212}{7}.\end{array}$$

Our area is $\frac{1}{2}\cdot 90\cdot \frac{212}{7}=1362\frac{6}{7}\approx \boxed{1363}.$