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Difference between revisions of "1984 AIME Problems/Problem 9"
(there's clearly a couple of typos in the prob, + solution, box)
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== Problem ==
 
== Problem ==
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In tetrahedron <math>\displaystyle ABCD</math>, edge <math>\displaystyle ABC</math> has length 3 cm. The area of face <math>\displaystyle AMC</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>\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>.
  
 
== Solution ==
 
== Solution ==
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{{solution}}
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{{image}}
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Position face <math>ABC</math> on the bottom. Since <math>[\triangle ABD] = 12 = \frac{1}{2} AB \cdot h_{ABD}</math>, we find that <math>h_{ABD} = 8</math>. The height of <math>ABD</math> forms a <math>30-60-90</math> with the height of the tetrahedron, so <math>h = \frac{1}{2} 8 = 4</math>. The volume of the tetrahedron is thus <math>\frac{1}{3}Bh = \frac{1}{3} 15 \cdot 4 = 020</math>.
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== See also ==
 
== See also ==
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* [[1984 AIME Problems/Problem 8 | Previous problem]]
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{{AIME box|year=1984|num-b=8|num-a=10}}
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* [[1984 AIME Problems/Problem 10 | Next problem]]
 
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* [[1984 AIME Problems]]
 

Revision as of 21:44, 5 March 2007

Problem

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

Solution

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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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