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Difference between revisions of "1984 AIME Problems/Problem 9"

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== Problem ==
 
== Problem ==
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In [[tetrahedron]] <math>\displaystyle ABCD</math>, [[edge]] <math>\displaystyle AB</math> has length 3 cm. The area of [[face]] <math>\displaystyle ABC</math> is <math>\displaystyle 15\mbox{cm}^2</math> and the area of face <math>\displaystyle ABD</math> is <math>\displaystyle 12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\displaystyle \mbox{cm}^3</math>.
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In [[tetrahedron]] <math>ABCD</math>, [[edge]] <math>AB</math> has length 3 cm. The area of [[face]] <math>ABC</math> is <math>15\mbox{cm}^2</math> and the area of face <math>ABD</math> is <math>12 \mbox { cm}^2</math>. These two faces meet each other at a <math>30^\circ</math> angle. Find the [[volume]] of the tetrahedron in <math>\mbox{cm}^3</math>.
  
 
== Solution ==
 
== Solution ==
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* [[American Invitational Mathematics Examination]]
 
* [[American Invitational Mathematics Examination]]
 
* [[Mathematics competition resources]]
 
* [[Mathematics competition resources]]
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[[Category:Intermediate Geometry Problems]]

Revision as of 08:37, 18 October 2007

Problem

In tetrahedron $ABCD$, edge $AB$ has length 3 cm. The area of face $ABC$ is $15\mbox{cm}^2$ and the area of face $ABD$ is $12 \mbox { cm}^2$. These two faces meet each other at a $30^\circ$ angle. Find the volume of the tetrahedron in $\mbox{cm}^3$.

Solution

1984 AIME-9.png

Position face $ABC$ on the bottom. Since $[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}$, we find that $h_{ABD} = 8$. The height of $ABD$ forms a $30-60-90$ with the height of the tetrahedron, so $h = \frac{1}{2} 8 = 4$. The volume of the tetrahedron is thus $\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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