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AMC 10 Geometry Math Olympiad USA Math Olympiad

Surface area of Cube Problem | AMC-10A, 2007 | Problem 21

Try this beautiful problem from Geometry, based on Cube from AMC-10A, 2007. You may use sequential hints to solve the problem

Try this beautiful problem from Geometry based on Surface area of a cube.

Surface area of cube – AMC-10A, 2007- Problem 21


A sphere is inscribed in a cube that has a surface area of \(24\) square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?

  • \(3\)
  • \(4\)
  • \(8\)
  • \(4\)

Key Concepts


Geometry

Cube

square

Check the Answer


Answer: \(8\)

AMC-10A (2007) Problem 21

Pre College Mathematics

Try with Hints


Surface area of Cube - problem

We have to find out the surface area of the inner cube.but to find out the surface area of inner cube the side length is require.but we don’t know the side length of the inner cube.but if you see the above diagram you must notice that  two opposite vertices of the cube are on opposite faces of the larger cube.Therefore two opposite vertices of the cube are on opposite faces of the larger cube.Thus the diagonal of the smaller cube is the side length of the outer square…..

Can you now finish the problem ……….

surface area of inner cube

The area of each face of the outer cube is \(\frac {24}{6} = 4\).Therefore  the edge length of the outer cube is \(2\). so the diagonal of the inner cube is \(2\).let the side length of inner cube is \(x\)

Therefore diagonal =\(\sqrt {x^2 +({\sqrt 2}x)^2}=2\) \(\Rightarrow x=\frac{2}{\sqrt 3}\)

can you finish the problem……..

Therefore surface area of the inner cube is \(6x^2\)=\(6 \times (\frac{2}{\sqrt 3})^2\)=\(8\)

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