Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.
Faces ABC and BCD of tetrahedron ABCD meet at an angle of 30,The area of face ABC=120, the area of face BCD is 80, BC=10. Find volume of tetrahedron.
Area
Volume
Tetrahedron
But try the problem first...
Answer: is 320.
AIME I, 1992, Question 6
Coordinate Geometry by Loney
First hint
Area BCD=80=\(\frac{1}{2} \times {10} \times {16}\),
where the perpendicular from D to BC has length 16.
Second Hint
The perpendicular from D to ABC is 16sin30=8
[ since sin30=\(\frac{perpendicular}{hypotenuse}\) then height = perpendicular=hypotenuse \(\times\) sin30 ]
Final Step
or, Volume=\(\frac{1}{3} \times 8 \times 120\)=320.
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