Triangle $PAB$ and square $ABCD$ are in [perpendicular](https://artofproblemsolving.com/wiki/index.php/Perpendicular) planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$? 

$\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$
Solution 1
Since the two planes are perpendicular, it follows that $\triangle PAD$ is a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle). Thus, $PD = \sqrt{PA^2 + AD^2} = \sqrt{PA^2 + AB^2} = \sqrt{34}$, which is option $\boxed{\text{B}}$.

Solution 2
Place the points on a coordinate grid, and let the $xy$ plane (where $z=0$) contain triangle $PAB$.  Square $ABCD$ will have sides that are vertical.
Place point $P$ at $(0,0,0)$, and place $PA$ on the x-axis so that $A(3,0,0)$, and thus $PA = 3$.
Place $PB$ on the y-axis so that $B(0,4,0)$, and thus $PB = 4$.  This makes $AB = 5$, as it is the hypotenuse of a 3-4-5 right triangle (with the right angle being formed by the x and y axes).  This is a clean use of the fact that $\triangle PAB$ is a right triangle.
Since $AB$ is one side of $\square ABCD$ with length $5$, $BC = 5$ as well.  Since $BC \perp AB$, and $BC$ is also perpendicular to the $xy$ plane, $BC$ must run stright up and down.  WLOG pick the up direction, and since $BC = 5$, we travel $5$ units up to $C(0,4,5)$.  Similarly, we travel $5$ units up from $A(3,0,0)$ to reach $D(3,0,5)$.
We now have coordinates for $P$ and $D$.  The distance is $\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \sqrt{34}$, which is option $\boxed{\text{B}}$.