The Height is There

Because triangle $AEC$ is right-angle, we can solve for the hypotenuse via the Pythagorean Theorem.

$3^{2}+4^{2}=c^{2}$

$25=c^{2}$

$5=c$

Now, we need to find the length of line $\overline{CD}$ .

We observe that the area of rectangle $ABDC = (\overline{AC})(\overline{CD})$ .

We established that $\overline{AC}=c=5$ , and therefore

$\frac{(5)(\overline{CD})}{2} = 6$

because $\overline{CD}$ can be considered a height if $\overline{AC}$ is the base.

By isolating the variable, we get

$\overline{CD}=\frac{12}{5}$

$ABDC=(\overline{AC})(\overline{CD})$

Plugging our values in...

$ABDC=(5)(\frac{12}{5})$

We find the area of rectangle $ABDC=\boxed{12}$ .

Shortcut method:

$ABDC=ab=(3)(4)=\boxed{12}$ .