Find the length of the longest rod that can be put in the room of dimensions $ 10 \mathrm{~cm} \times 6 \mathrm{~cm} $ $ \times 4 \mathrm{~cm} . \quad $

Given:

The dimensions of the room are \( 10 \mathrm{~cm} \times 6 \mathrm{~cm} \) \( \times 4 \mathrm{~cm} . \quad \)

To do:

We have to find the length of the longest rod that can be put in the room.

Solution:

The length of the longest rod that can be put in the room is the length of the diagonal formed by the cuboid(room).
Length$l=10\ cm$

Breadth$b=6\ cm$

Height$h=4\ cm$

Diagonal of a cuboid of length $l$, breadth $b$ and height $h$ is $\sqrt{l^2+b^2+h^2}$.

Therefore,

Length of the diagonal of the room$=\sqrt{10^2+6^2+4^2}\ cm$

$=\sqrt{100+36+16}\ cm$

$=\sqrt{152}\ cm$

$=\sqrt{4\times38}\ cm$

$=2\sqrt{38}\ cm$
 The length of the longest rod that can be put in the room is $2\sqrt{38}\ cm$.

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Updated on: 10-Oct-2022

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