Difference between revisions of "1996 AHSME Problems/Problem 9"
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==Problem== | ==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> | <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== | ||
+ | === 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. | 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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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>. | 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== | ==See also== | ||
{{AHSME box|year=1996|num-b=8|num-a=10}} | {{AHSME box|year=1996|num-b=8|num-a=10}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Revision as of 21:32, 18 August 2011
Problem
Triangle and square are in perpendicular planes. Given that and , what is ?
Contents
[hide]Solution
Solution 1
Since the two planes are perpendicular, it follows that is a right triangle. Thus, , which is option .
Solution 2
Place the points on a coordinate grid, and let the plane (where ) contain triangle . Square will have sides that are vertical.
Place point at , and place on the x-axis so that , and thus .
Place on the y-axis so that , and thus . This makes , 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 is a right triangle.
Since is one side of with length , as well. Since , and is also perpendicular to the plane, must run stright up and down. WLOG pick the up direction, and since , we travel units up to . Similarly, we travel units up from to reach .
We now have coordinates for and . The distance is , which is option .
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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All AHSME Problems and Solutions |