# 1996 AHSME Problems/Problem 9

## Problem

Triangle $PAB$ and square $ABCD$ are in perpendicular planes. Given that $PA = 3, PB = 4$ and $AB = 5$, what is $PD$? $[asy] real r=sqrt(2)/2; draw(origin--(8,0)--(8,-1)--(0,-1)--cycle); draw(origin--(8,0)--(8+r, r)--(r,r)--cycle); filldraw(origin--(-6*r, -6*r)--(8-6*r, -6*r)--(8, 0)--cycle, white, black); filldraw(origin--(8,0)--(8,6)--(0,6)--cycle, white, black); pair A=(6,0), B=(2,0), C=(2,4), D=(6,4), P=B+1*dir(-65); draw(A--P--B--C--D--cycle); dot(A^^B^^C^^D^^P); label("A", A, dir((4,2)--A)); label("B", B, dir((4,2)--B)); label("C", C, dir((4,2)--C)); label("D", D, dir((4,2)--D)); label("P", P, dir((4,2)--P)); [/asy]$ $\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$

## Solution

### Solution 1

Since the two planes are perpendicular, it follows that $\triangle PAD$ is a 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 $\boxed{\text{B}}$.

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