Difference between revisions of "1992 AIME Problems/Problem 7"
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== Solution == | == Solution == | ||
− | Since the area <math>BCD=80=\frac{1}{2}\cdot10\cdot16</math>, the perpendicular from D to BC has length | + | Since the area <math>BCD=80=\frac{1}{2}\cdot10\cdot16</math>, the perpendicular from <math>D</math> to <math>BC</math> has length <math>16</math>. |
− | The perpendicular from D to ABC is <math>16 \cdot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>. | + | The perpendicular from <math>D</math> to <math>ABC</math> is <math>16 \cdot \sin 30^\circ=8</math>. Therefore, the volume is <math>\frac{8\cdot120}{3}=\boxed{320}</math>. |
== See also == | == See also == |
Latest revision as of 16:40, 14 March 2017
Problem
Faces and of tetrahedron meet at an angle of . The area of face is , the area of face is , and . Find the volume of the tetrahedron.
Solution
Since the area , the perpendicular from to has length .
The perpendicular from to is . Therefore, the volume is .
See also
1992 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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