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 $PD$ ?

$\text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8$

Solution

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 $\boxed{B}$

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

Page

Toolbox

Search

1996 AHSME Problems/Problem 9

Problem

Solution

See also