What is the length of the greatest rod that can be placed in a room whose length is $10$ meter, breadth $8$ meter and height $6$ meter?
Given that $\sqrt{2} = 1.42$

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Solution

Length of the greatest (i.e. longest ) rod which can be placed in a room
= length of the diagonal
$= \sqrt{l^{2} + b^{2} + b^{2}}$
$= \sqrt{10^{2} + 8^{2} + 6^{2}}$
$= \sqrt{200}$
$= 10 \sqrt{2}$ meter
$= 10 \times 1.42 m$
$= 14.2 m$

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