# 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>. | ||

+ | |||

+ | 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>. | ||

+ | |||

+ | 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 23:16, 23 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 .