Difference between revisions of "1983 AIME Problems/Problem 11"
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<math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math> | <math>V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}</math> | ||
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+ | == Solution 3 == | ||
+ | We can also find the volume by integrating horizontal cross-sections of the solid. As in solution 1, we can find the height of the solid is <math>6</math>; thus, we will integrate with respect to height from <math>0</math> to <math>6</math>, noting that each cross section of height <math>dh</math> is a rectangle. The problem becomes <math>\int_0^h(wl)dh</math>, where <math>w</math> is the width of the rectangle and <math>l</math> is the length. We can express <math>w</math> in terms of <math>h</math> as <math>w=6\sqrt{2}-\sqrt{2}h</math> since it decreases linearly with respect to <math>h</math>, and <math>l=6\sqrt{2}+\sqrt{2}h</math> since the length increases linearly with respect to <math>h</math>. Now we solve:<cmath>\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)d=\int_0^6(72-2h^2)dh=72(6)-2(\frac{1}{3})(6^3)=\boxed{288}</cmath> | ||
== See Also == | == See Also == |
Revision as of 10:50, 13 June 2018
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 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 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 integrating horizontal cross-sections of the solid. As in solution 1, we can find 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 problem becomes , 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 the length increases linearly with respect to . Now we solve:
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 |