1983 AIME Problems/Problem 11
Contents
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?
Solutions
Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of to . The hypotenuse of the triangle formed is the median of equilateral triangle , and one of the legs is . We apply the Pythagorean Theorem to deduce that the height is .
Next, we complete t he figure into a triangular prism, and find its volume, which is .
Now, we subtract off the two extra pyramids that we included, whose combined volume is .
Thus, our answer is .
Solution 2
Extend and to meet at , and and to meet at . Now, we have a regular tetrahedron , which by symmetry 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:
.
Solution 3
We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is ; thus, we will integrate with respect to height from to , noting that each cross section of height is a rectangle. The volume is then , where is the width of the rectangle and is the length. We can express in terms of as since it decreases linearly with respect to , and since it similarly increases linearly with respect to . Now we solve:.
Solution 4
Draw an altitude from a vertex of the square base to the top edge. By using triangle ratios, we obtain that the altitude has a length of , and that little portion that hangs out has a length of . This is a triangular pyramid with a base of , and a height of . Since there are two of these, we can compute the sum of the volumes of these two to be . Now we are left with a triangular prism with a base of dimensions and a height of . We can compute the volume of this to be 216, and thus our answer is .
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Solution 5
From solution 1, the height of the solid is . Construct a triangular prism with base ABCD, and with the height of the solid. The volume of this triangular prism is Notice that the solid is symetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a regular tetrahedron. This tetrahedron has sidelength of and since the formula for the volume of a regular tetrahedron is the volume of this tetrahedron is . .
~skibidi solver
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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