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?
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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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[[Image:1983Number11.JPG]]
 
[[Image:1983Number11.JPG]]
 
 
== 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>.
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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 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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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>.
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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}}
* [[AIME Problems and Solutions]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 12:07, 25 April 2008

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? 1983Number11.JPG

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 $432-144=\boxed{288}$.

See also

1983 AIME (ProblemsAnswer KeyResources)
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
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