The distance

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

#Geometry

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Tanishq Varshney - 6 years ago

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

Tanishq Varshney - 6 years ago

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

vishnu c - 6 years ago

@vishnu c @Nishant Rai

Tanishq Varshney - 6 years ago

The ans is 2 $\sqrt{\frac{8}{3}}$

Naman Kapoor - 6 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$