Difference between revisions of "1983 AIME Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | The solid shown has a square base of side length <math>s</math>. The upper edge is parallel to the base and has length <math>2s</math>. All edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid? | + | The solid shown has a [[square]] base of side length <math>s</math>. The upper edge is [[parallel]] to the base and has length <math>2s</math>. All edges have length <math>s</math>. Given that <math>s=6\sqrt{2}</math>, what is the volume of the solid? |
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| + | </asy></center> -->[[Category:Asymptote needed]] | ||
[[Image:1983Number11.JPG]] | [[Image:1983Number11.JPG]] | ||
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== Solution == | == Solution == | ||
| − | 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 | + | 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>. |
Next, we complete the figure into a triangular prism, and find the area, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>. | Next, we complete the figure into a triangular prism, and find the area, which is <math>\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432</math>. | ||
| − | Now, we subtract off the two extra | + | Now, we subtract off the two extra [[pyramid]]s that we included, whose combined area is <math>2\cdot \left( \frac{12\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144</math>. |
| − | Thus, our answer is <math>432-144=288</math>. | + | Thus, our answer is <math>432-144=\boxed{288}</math>. |
== See also == | == See also == | ||
{{AIME box|year=1983|num-b=10|num-a=12}} | {{AIME box|year=1983|num-b=10|num-a=12}} | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 11:07, 25 April 2008
Problem
The solid shown has a square base of side length 
. The upper edge is parallel to the base and has length 
. All edges have length . Given that 
, what is the volume of the solid?
Solution
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 
.
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 | ||
