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## 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

Area BCD=80=$\frac{1}{2} \times {10} \times {16}$,

where the perpendicular from D to BC has length 16.

The perpendicular from D to ABC is 16sin30=8

[ since sin30=$\frac{perpendicular}{hypotenuse}$ then height = perpendicular=hypotenuse $\times$ sin30 ]

or, Volume=$\frac{1}{3} \times 8 \times 120$=320.

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## Tetrahedron Problem | AMC-10A, 2011 | Problem 24

Try this beautiful problem from Geometry based on Tetrahedron.

## Tetrahedron Problem – AMC-10A, 2011- Problem 24

Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?

• $4\sqrt 5+3\sqrt 2$
• $4\sqrt 5+3\sqrt 2$
• $\frac{1}{6}$
• $7\sqrt 3+4\sqrt 2$
• $\frac{1}{8}$

### Key Concepts

Geometry

Tetrahedron

Pythagoras

Answer: $\frac{1}{6}$

AMC-10A (2011) Problem 24

Pre College Mathematics

## Try with Hints

We know that if we split a regular tetrahedron then it will split into eight tetrahedra that have lengths of $\frac{1}{2}$. The volume of a regular tetrahedron can be found using base area and height. Let us assume that the side length of the above tetrahedron is 1, its base area is $\frac{\sqrt{3}}{4}$, and its height can be obtaine using Pythagoras’ Theorem.Can you find out the height …..

can you finish the problem……..

Therefore the height will be $\sqrt{1^2-(\frac{\sqrt 3}{3})^2}$=$\frac{\sqrt2}{\sqrt 3}$ And the volume will be

$\frac{1}{3} .\frac{\sqrt 3}{4}.\frac{\sqrt 2}{\sqrt 3$=$\frac{\sqrt 2}{12}$

Now the actual side length of the Tetrahedron=$\sqrt 2$.Therefore the actual volume is $\frac{\sqrt 2}{12}.{\sqrt 2}^3=\frac{1}{3}$

can you finish the problem……..

There are eight small tetrahedron, the four tetrahedra on the corners of the large tetrahedra are not inside the other large tetrahedra. Thus, $\frac{4}{8}=\frac{1}{2}$ of the large tetrahedra will not be inside the other large tetrahedra.

Therefore  the volume of the region formed by the intersection of the tetrahedron is $\frac{1}{3} \times \frac{1}{3}$=$\frac{1}{6}$