8. (15 points) In quadrilateral $A B C D$, $A B=B C=9$ cm, $A D-D C=8$ cm, $A B$ is perpendicular to $B C$, $A D$ is perpendicular to $D C$. Then the area of quadrilateral $A B C D$ is $\qquad$ square cm.
Solution: According to the analysis, as shown in the figure, let $CE = x$, then $AC = x + 8$.
In the right triangle $\triangle ABC$, by the Pythagorean theorem, we have: $AB^2 + BC^2 = AC^2$.
In the right triangle $\triangle ADC$, $AD^2 + DC^2 = AC^2$.
That is: $AB^2 + BC^2 = AD^2 + DC^2$
$$
9^2 + 9^2 = (x + 8)^2 + x^2
$$

We get: $x^2 + 8x = 49$
$\therefore x(x + 8) = 49$
$S_{\triangle ADC} = \frac{1}{2} \times AD \times CD = \frac{1}{2} x(x + 8) = \frac{1}{2} \times 49 = \frac{49}{2}$
$S_{\triangle ABC} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 9 \times 9 = \frac{81}{2}$
$S_{\text{quadrilateral } ABCD} = S_{\triangle ADC} + S_{\triangle ABC} = \frac{49}{2} + \frac{81}{2} = 65$
Therefore, the answer is: 65