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

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 $\boxed{\text{B}}$ .

1996 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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1996 AHSME Problems/Problem 9

Problem

Contents

Solution

Solution 1

Solution 2

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