Practicing Success
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What is the maximum distance between any two points of a cube of side $l$ units? |
$(\sqrt{2}+1)l\,units$ $\sqrt{2}l\,units$ $\sqrt{3}l\,units$ $2\, l\, units$ |
$\sqrt{3}l\,units$ |
The maximum distance between any two points of a cube of side $‘l’$ units is the diagonal. Hence, diagonal of cube = $\sqrt{l^2+l^2+l^2}=\sqrt{3l^2}=\sqrt{3}\,l\,units$ |
