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April 20, 2020

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

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




Check the Answer

Answer: \(\frac{1}{6}\)

AMC-10A (2006) Problem 24

Pre College Mathematics

Try with Hints

Octahedron Problem figure

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 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 ..........


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........

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

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