Difference between revisions of "1996 AHSME Problems/Problem 9"

Revision as of 21:32, 18 August 2011

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

@@ Line 1: / Line 1: @@
 ==Problem==
-Triangle <math>PAB</math> and square <math>ABCD</math> are in perpendicular planes. Given that <math>PA = 3, PB = 4</math> and <math>AB = 5</math>, what is <math>PD</math>?
+Triangle <math>PAB</math> and square <math>ABCD</math> are in [[perpendicular]] planes. Given that <math>PA = 3, PB = 4</math> and <math>AB = 5</math>, what is <math>PD</math>?
 <math> \text{(A)}\ 5\qquad\text{(B)}\ \sqrt{34} \qquad\text{(C)}\ \sqrt{41}\qquad\text{(D)}\ 2\sqrt{13}\qquad\text{(E)}\ 8 </math>
+__TOC__
 ==Solution==
+=== Solution 1 ===
+Since the two planes are perpendicular, it follows that <math>\triangle PAD</math> is a [[right triangle]]. Thus, <math>PD = \sqrt{PA^2 + AD^2} = \sqrt{PA^2 + AB^2} = \sqrt{34}</math>, which is option <math>\boxed{\text{B}}</math>.
+=== Solution 2 ===
 Place the points on a coordinate grid, and let the <math>xy</math> plane (where <math>z=0</math>) contain triangle <math>PAB</math>.  Square <math>ABCD</math> will have sides that are vertical.
@@ Line 15: / Line 19: @@
 Since <math>AB</math> is one side of <math>\square ABCD</math> with length <math>5</math>, <math>BC = 5</math> as well.  Since <math>BC \perp AB</math>, and <math>BC</math> is also perpendicular to the <math>xy</math> plane, <math>BC</math> must run stright up and down.  WLOG pick the up direction, and since <math>BC = 5</math>, we travel <math>5</math> units up to <math>C(0,4,5)</math>.  Similarly, we travel <math>5</math> units up from <math>A(3,0,0)</math> to reach <math>D(3,0,5)</math>.
-We now have coordinates for <math>P</math> and <math>D</math>.  The distance is <math>\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \sqrt{34}</math>, which is option <math>\boxed{B}</math>
+We now have coordinates for <math>P</math> and <math>D</math>.  The distance is <math>\sqrt{(5-0)^2 + (0-0)^2 + (3-0)^2} = \sqrt{34}</math>, which is option <math>\boxed{\text{B}}</math>.
 ==See also==
 {{AHSME box|year=1996|num-b=8|num-a=10}}
+[[Category:Introductory Geometry Problems]]
+[[Category:3D Geometry Problems]]

1996 AHSME (Problems • Answer Key • Resources)
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