Difference between revisions of "AoPS Wiki talk:Problem of the Day/June 11, 2011"
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MP("X",X,(-1,-1)); | MP("X",X,(-1,-1)); | ||
</asy> | </asy> | ||
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+ | If one were to find the volume of the pyramid, you would need to look for the area of the base and the height. | ||
+ | |||
+ | The area of the base is square <math>JKLM</math>. | ||
+ | |||
+ | <math>(\dfrac{1}{2})^3=\frac{1}{8}</math> | ||
+ | |||
+ | To find the height, you would have to apply the Pythagorean Theorem on a triangle with vertices that are the apex, point X, and a vertex of square JKLM. | ||
+ | This triangle however is identical to triangle <math>AEJ</math>. Thus, the height of the pyramid is equal to <math>\dfrac{1}{2}</math>. | ||
+ | |||
+ | Therefore, the volume of the pyramid is : | ||
+ | |||
+ | <math>\dfrac{1}{8}\times\dfrac{1}{2}\times\dfrac{1}{3}=\boxed{\dfrac{1}{48}}</math> |
Revision as of 21:24, 10 June 2011
Problem
AoPSWiki:Problem of the Day/June 11, 2011
Solution
This Problem of the Day needs a solution. If you have a solution for it, please help us out by adding it.
If one were to find the volume of the pyramid, you would need to look for the area of the base and the height.
The area of the base is square .
To find the height, you would have to apply the Pythagorean Theorem on a triangle with vertices that are the apex, point X, and a vertex of square JKLM.
This triangle however is identical to triangle . Thus, the height of the pyramid is equal to
.
Therefore, the volume of the pyramid is :