Tetrahedron Problem | AIME I, 1992 | Question 6

Content

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Tetrahedron.

Tetrahedron Problem – AIME I, 1992

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.

Source

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.