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

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

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

Log in to reply

@vishnu c @Nishant Rai

Log in to reply

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

Log in to reply