Difference between revisions of "1983 AIME Problems/Problem 11"
(→Problem) |
(→Solution 1) |
||
Line 27: | Line 27: | ||
triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); | triple A=(0,0,0),B=(s,0,0),C=(s,s,0),D=(0,s,0),E=(-s/2,s/2,6),F=(3*s/2,s/2,6); | ||
triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); | triple Aa=(E.x,0,0),Ba=(F.x,0,0),Ca=(F.x,s,0),Da=(E.x,s,0); | ||
− | + | draw(A--B--C--D--A--E--D); | |
− | + | draw(B--F--C); | |
− | + | draw(E--F); | |
− | + | draw(B--Ba--Ca--C,dashed+d); | |
+ | draw(A--Aa--Da--D,dashed+d); | ||
+ | draw(E--(E.x,E.y,0),dashed+l); | ||
+ | draw(F--(F.x,F.y,0),dashed+l); | ||
+ | draw(Aa--E--Da,dashed+d); | ||
+ | draw(Ba--F--Ca,dashed+d); | ||
+ | label("A",A); | ||
+ | label("B",B); | ||
+ | label("C",C); | ||
+ | label("D",D); | ||
+ | label("E",E,N); | ||
+ | label("F",F,N); | ||
+ | label("$12\sqrt{2}$",(E+F)/2,N); | ||
+ | label("$6\sqrt{2}$",(A+B)/2); | ||
+ | label("6",(3*s/2,s/2,3),ENE); | ||
</asy></center> | </asy></center> | ||
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>. |
Revision as of 16:03, 26 March 2011
Contents
Problem
The solid shown has a square base of side length . The upper edge is parallel to the base and has length . All other edges have length . Given that , what is the volume of the solid?
Solution 1
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 .
Solution 2
Extend and to meet at , and and to meet at . now, we have a regular tetrahedron , which has twice the volume of our original solid. This tetrahedron has side length . Using the formula for the volume of a regular tetrahedron, which is , where S is the side length of the tetrahedron, the volume of our original solid is:
See also
1983 AIME (Problems • Answer Key • Resources) | ||
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 |