Difference between revisions of "1983 AIME Problems/Problem 11"
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=== Solution 5 === | === Solution 5 === | ||
− | From solution 1, the height of the solid is 6. Construct a triangular prism with base | + | <cmath> From solution 1, the height of the solid is 6. Construct a triangular prism with base |
\[ | \[ | ||
\left( 6\sqrt{2} \right)^2 \times 6 \times \frac{1}{2} = 216. | \left( 6\sqrt{2} \right)^2 \times 6 \times \frac{1}{2} = 216. | ||
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\] | \] | ||
− | \] | + | \] </cmath> |
== See Also == | == See Also == |
Revision as of 02:24, 27 May 2024
Contents
[hide]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 .
pi_is_3.141
Solution 5
From solution 1, the height of the solid is 6. Construct a triangular prism with base \( ABCD \), and with the height of the solid. The volume of this rectangular prism is \[ \left( 6\sqrt{2} \right)^2 \times 6 \times \frac{1}{2} = 216. \] Notice that the solid is symmetrical, and if you remove the triangular prism described earlier and combine the two halves on each side of it, you will get a tetrahedron. This tetrahedron has a side length of \( 6\sqrt{2} \) and since the formula for the volume of a tetrahedron is \[ \frac{s^3}{6\sqrt{2}}, \] the volume of this tetrahedron is \[ \frac{(6\sqrt{2})^3}{6\sqrt{2}} = 72. \] Thus, the total volume is \[ 216 + 72 = \boxed{288}. \] \] (Error compiling LaTeX. Unknown error_msg)
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 |