Difference between revisions of "1996 AHSME Problems/Problem 9"
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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{ | + | 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> |
==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 20:35, 18 August 2011
Problem
Triangle and square are in perpendicular planes. Given that and , what is ?
Solution
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 |