# Tetrahedron Problem | AIME I, 1992 | Question 6

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.

- is 107
- is 320
- is 840
- cannot be determined from the given information

**Key Concepts**

Area

Volume

Tetrahedron

## Check the Answer

But try the problem first...

Answer: is 320.

AIME I, 1992, Question 6

Coordinate Geometry by Loney

## Try with Hints

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA