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# Octahedron Problem | AMC-10A, 2006 | Problem 24

Try this beautiful problem from Geometry: Octahedron AMC-10A, 2006. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Octahedron

## Octahedron – AMC-10A, 2006- Problem 24

Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?

• $\frac{1}{3}$
• $\frac{1}{6}$
• $\frac{3}{4}$

### Key Concepts

Geometry

Octahedron

pyramid

But try the problem first…

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

Source

AMC-10A (2006) Problem 24

Pre College Mathematics

## Try with Hints

First hint

Now we have to find out the volume of the octahedron.given that Centers of adjacent faces of a unit cube are joined to form a regular octahedron.so if we divide the octahedron two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals the we will get two pyramids .Now if we can find out the area of Pyramids them we can find out the volume of Octahedron.can you find out the volume of the Pyramids?

Can you now finish the problem ……….

Second Hint

Given that side length of a cube is $1$.Therefore length of all edges of the regular octahedron =$\frac{\sqrt 2}{2}$.Now the base of the pyramids is a square are will be $(\frac{\sqrt 2}{2})^2$=$\frac{1}{2}$.Now clearly the height of the pyramid is half the height of the cube, i.e $\frac{1}{2}$.So volume of the Pyramid will be $\frac{1}{3} \times$ (Base area) $\times$ (height)=$\frac{1}{3} \times \frac{1}{2} \times \frac{1}{2}$=$\frac{1}{12}$

can you finish the problem……..

Final Step

Therefore the aree of the octahedron= 2 $\times$ area of Pyramid= 2 $\times \frac{1}{12}$=$\frac{1}{6}$

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