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

(Created page with "==See also== {{AHSME box|year=1996|num-b=8|num-a=10}}")
 
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==Problem==
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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>?
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<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>
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==Solution==
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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.
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Place point <math>P</math> at <math>(0,0,0)</math>, and place <math>PA</math> on the x-axis so that <math>A(3,0,0)</math>, and thus <math>PA = 3</math>.
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Place <math>PB</math> on the y-axis so that <math>B(0,4,0)</math>, and thus <math>PB = 4</math>.  This makes <math>AB = 5</math>, 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 <math>\triangle PAB</math> is a right triangle.
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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>.
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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{D}</math>
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=8|num-a=10}}
 
{{AHSME box|year=1996|num-b=8|num-a=10}}

Revision as of 20:33, 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

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 $\boxed{D}$

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
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All AHSME Problems and Solutions
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