jmc

The center of the circle, $O$, is the midpoint of the chord $AB$ (the diameter of the circle). Since we are told that $CD$ is parallel to $AB$, if we draw a line that is perpendicular to $AB$, it will be perpendicular to $CD$ as well. Now let's draw a segment from $O$ to the midpoint of the chord $CD$, which we will call $X$, and another segment from $O$ to $D$. Now we have right triangle $OXD$ as shown: We are told that chord $CD$ is 8 units long. Since $X$ is the midpoint of chord $CD$, both $CX$ and $XD$ must be 4 units long. We are also told that circle $O$ has a radius of 6 units. This means that $OD$ must be 6 units long. Because we have a right triangle, we can use the Pythagorean Theorem to find the length of $OX$. We get $$\begin{array}{rl}O{X}^2+X{D}^2& =O{D}^2\\ O{X}^2& =O{D}^2-X{D}^2\\ OX& =\sqrt{O{D}^2-X{D}^2}\\ OX& =\sqrt{6^2-4^2}\\ OX& =\sqrt{20}.\end{array}$$ Now let's draw a segment from $D$ to a point $Y$ on segment $KA$ that is perpendicular to both $CD$ and $KA$. We get $DY$, drawn in red in the following diagram: Since $DY$ forms right triangle $DYO$, which is congruent to $△OXD$, we get that $DY$ is $\sqrt{20}$ units long. Now we can use the formula for a triangle, $\text{mbox}area=\frac{1}{2}\text{mbox}base\cdot \text{mbox}height$ to get the area of $△KDC$. We get $$\begin{array}{rl}\text{mbox}area& =\frac{1}{2}\cdot 8\cdot \sqrt{20}\\ & =4\cdot \sqrt{20}\\ & =4\cdot 2\sqrt{5}\\ & =\boxed{8\sqrt{5}}.\end{array}$$