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1983 AIME Problems/Problem 11: Difference between revisions

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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>.
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* [[1983 AIME Problems/Problem 10|Previous Problem]]
* [[1983 AIME Problems/Problem 12|Next Problem]]
* [[1983 AIME Problems|Back to Exam]]


== See also ==
== See also ==
* [[1983 AIME Problems]]
* [[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 $s$. The upper edge is parallel to the base and has length $2s$. All edges have length $s$. Given that $s=6\sqrt{2}$, 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 $AD$ to $EF$. The hypotenuse of the triangle is the median of equilateral triangle $ADE$ one of the legs is $3\sqrt{2}$. We apply the pythagorean theorem to find that the height is equal to $6$.

Next, we complete the figure into a triangular prism, and find the area, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$.

Now, we subtract off the two extra pyramids that we included, whose combined area is $2\cdot \left( \frac{12\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$.

Thus, our answer is $432-144=288$.

See also

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