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? | ||
| + | [img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=791&sid=cfd5dae222dd7b8944719b56de7b8bf7[/img] | ||
| + | {{img}} | ||
== 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 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>. | ||
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| + | Now, we subtract off the two extra pyramids 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>. | ||
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| + | Thus, our answer is <math>432-144=288</math>. | ||
| + | |||
| + | ---- | ||
| + | |||
| + | * [[1983 AIME Problems/Problem 10|Previous Problem]] | ||
| + | * [[1983 AIME Problems/Problem 12|Next Problem]] | ||
| + | * [[1983 AIME Problems|Back to Exam]] | ||
== See also == | == See also == | ||
| − | * [[ | + | * [[AIME Problems and Solutions]] |
| + | * [[American Invitational Mathematics Examination]] | ||
| + | * [[Mathematics competition resources]] | ||
| + | |||
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 00:16, 24 July 2006
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?
[img]http://www.artofproblemsolving.com/Forum/album_pic.php?pic_id=791&sid=cfd5dae222dd7b8944719b56de7b8bf7[/img]
Template:Img
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 
.
