1996 AHSME Problems/Problem 9
Problem
Triangle and square are in perpendicular planes. Given that and , what is ?
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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