Difference between revisions of "1983 AIME Problems/Problem 11"
(→Solution 1) |
m |
||
Line 18: | Line 18: | ||
</asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --> | </asy></center> <!-- Asymptote replacement for Image:1983Number11.JPG by bpms --> | ||
− | == Solution 1 == | + | == Solution == |
+ | === Solution 1 === | ||
First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of <math>AD</math> to <math>EF</math>. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] <math>ADE</math> one of the legs is <math>3\sqrt{2}</math>. We apply the [[Pythagorean Theorem]] to find that the height is equal to <math>6</math>. | First, we find the height of the figure by drawing a [[perpendicular]] from the midpoint of <math>AD</math> to <math>EF</math>. The [[hypotenuse]] of the triangle is the [[median]] of [[equilateral triangle]] <math>ADE</math> one of the legs is <math>3\sqrt{2}</math>. We apply the [[Pythagorean Theorem]] to find that the height is equal to <math>6</math>. | ||
<center><asy> | <center><asy> | ||
Line 52: | Line 53: | ||
Thus, our answer is <math>432-144=\boxed{288}</math>. | Thus, our answer is <math>432-144=\boxed{288}</math>. | ||
− | == Solution 2 == | + | === Solution 2 === |
Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. now, we have a regular tetrahedron <math>EFGH</math>, which has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is: | Extend <math>EA</math> and <math>FB</math> to meet at <math>G</math>, and <math>ED</math> and <math>FC</math> to meet at <math>H</math>. now, we have a regular tetrahedron <math>EFGH</math>, which has twice the volume of our original solid. This tetrahedron has side length <math>2s = 12\sqrt{2}</math>. Using the formula for the volume of a regular tetrahedron, which is <math>V = \frac{\sqrt{2}S^3}{12}</math>, where S is the side length of the tetrahedron, the volume of our original solid is: | ||
<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> | ||
− | == See | + | == See Also == |
{{AIME box|year=1983|num-b=10|num-a=12}} | {{AIME box|year=1983|num-b=10|num-a=12}} | ||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 07:02, 16 April 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 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 area, 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 |