# 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

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.

# Largest Area of Triangle | AIME I, 1992 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Largest Area of Triangle.

## Area of Triangle - AIME I, 1992

Triangle ABC has AB=9 and BC:AC=40:41, find the largest area that this triangle can have.

• is 107
• is 820
• is 840
• cannot be determined from the given information

### Key Concepts

Ratio

Area

Triangle

AIME I, 1992, Question 13

Coordinate Geometry by Loney

## Try with Hints

First hint

Let the three sides be 9, 40x, 41x

area = $$\frac{1}{4}\sqrt{(81^2-81x^2)(81x^2-1)} \leq \frac{1}{4}\frac{81^2-1}{2}$$

Second Hint

or, $$\frac{1}{4}\frac{81^2-1}{2}=\frac{1}{8}(81-1)(81+1)$$

Final Step

=(10)(82)

=820.