Difference between revisions of "1983 AIME Problems/Problem 11"

Line 11: Line 11:
Thus, our answer is <math>432-144=288</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 box|year=1983|num-b=10|num-a=12}}
* [[AIME Problems and Solutions]]
* [[AIME Problems and Solutions]]
* [[American Invitational Mathematics Examination]]
* [[American Invitational Mathematics Examination]]

Revision as of 14:12, 6 May 2007


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


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

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
Invalid username
Login to AoPS