Difference between revisions of "1983 AIME Problems/Problem 11"
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label("6",(3*s/2,s/2,3),ENE); | label("6",(3*s/2,s/2,3),ENE); | ||
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− | Next, we complete the figure into a triangular prism, and find the | + | Next, we complete the figure into a triangular prism, and find the volume, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>. |
Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is <math>2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>. | Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is <math>2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>. |
Revision as of 18:47, 13 June 2012
Problem
The solid shown has a square base of side length . The upper edge is parallel to the base and has length . All other edges have length . Given that , what is the volume of the solid?
Solution
Solution 1
First, we find the height of the figure by drawing a perpendicular from the midpoint of to . The hypotenuse of the triangle is the median of equilateral triangle , and one of the legs is . We apply the Pythagorean Theorem to find that the height is equal to .
Next, we complete the figure into a triangular prism, and find the volume, which is .
Now, we subtract off the two extra pyramids that we included, whose combined area is .
Thus, our answer is .
Solution 2
Extend and to meet at , and and to meet at . now, we have a regular tetrahedron , which has twice the volume of our original solid. This tetrahedron has side length . Using the formula for the volume of a regular tetrahedron, which is , where S is the side length of the tetrahedron, the volume of our original solid is:
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |