$\text{ABCD}$ is a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle), $\text{D}$ is the center of the [circle](https://artofproblemsolving.com/wiki/index.php/Circle), and $\text{B}$ is on the circle.  If $\text{AD}=4$ and $\text{CD}=3$, then the [area](https://artofproblemsolving.com/wiki/index.php/Area) of the shaded region is between

$\text{(A)}\ 4\text{ and }5 \qquad \text{(B)}\ 5\text{ and }6 \qquad \text{(C)}\ 6\text{ and }7 \qquad \text{(D)}\ 7\text{ and }8 \qquad \text{(E)}\ 8\text{ and }9$
The area of the shaded region is equal to the area of the quarter circle with the area of the rectangle taken away.  The area of the rectangle is $4\cdot 3=12$, so we just need the quarter circle.
Applying the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem) to $\triangle ADC$, we have \[(AC)^2=4^2+3^2\Rightarrow AC=5\] Since $ABCD$ is a rectangle, \[BD=AC=5\]
Clearly $BD$ is a [radius](https://artofproblemsolving.com/wiki/index.php/Radius) of the circle, so the area of the whole circle is $5^2\pi =25\pi$ and the area of the quarter circle is $\frac{25\pi }{4}$.
Finally, the shaded region is \[\frac{25\pi }{4}-12 \approx 7.6\] so the answer is $\boxed{\text{D}}$