Imagine in the same arrangement , all spheres are of same radius \(2~cm\) and kept on a horizontal plane, the distance between the top of the sphere kept above the three spheres from the plane is

A tetrahedron is formed of side \(4~cm\). The height of tetrahedron from its base to the centre of sphere is \( 4\frac{\sqrt{6}}{3}\), and \(2~cm \) below the base lies the plane, and \(2~cm\) above the centre is the top, did i do correctly

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TopNewest@Raghav Vaidyanathan , i got my answer as \(4+4\frac{\sqrt{6}}{3}\)

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A tetrahedron is formed of side \(4~cm\). The height of tetrahedron from its base to the centre of sphere is \( 4\frac{\sqrt{6}}{3}\), and \(2~cm \) below the base lies the plane, and \(2~cm\) above the centre is the top, did i do correctly

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Looks like it's correct to me. I checked it right now.

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@vishnu c @Nishant Rai

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The ans is 2\( \sqrt{\frac{8}{3}}\)

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