Question

The length, breadth and height of a rectangular box are $8$ cm, $6$ cm and $2$ cm respectively. Find the maximum length of a pencil that can be kept in this box.

Answer 1

Hint: For finding the maximum length of the pencil, we will use the formula of diagonal of a cuboid that is $\sqrt{lengt{{h}^{2}}+breadt{{h}^{2}}+heigh{{t}^{2}}}$ because rectangular box is in a shape of cuboid and the maximum length in a cuboid is its diagonal. After substituting the respective values in the formula, we will simplify the expression by using proper mathematical operations.

Complete step-by-step solution:
Since, the shape of the rectangular box is cuboid and diagonal length is the maximum length in it.
So, the maximum length of the pencil is equal to the length of the diagonal of the cuboid or rectangular box.
Here, we will use the formula of length of diagonal as:
$\Rightarrow diagonal=\sqrt{lengt{{h}^{2}}+breadt{{h}^{2}}+heigh{{t}^{2}}}$
Now, we will substitute $8$ for length, $6$ for breadth and $2$ for height in the above formula of diagonal as:
$\Rightarrow diagonal=\sqrt{{{8}^{2}}+{{6}^{2}}+{{2}^{2}}}$
After squaring $8$, \[6\] and $2$, we will have $64$, \[36\] and $4$ respectively and we will substitute these squares in the above step as:
$\Rightarrow diagonal=\sqrt{64+36+4}$
Now, we will add \[64,36\] and $4$. So, we will substitute $104$ in the place of \[64,36\] and $4$ in the above step as:
$\Rightarrow diagonal=\sqrt{104}$
Now, we will factorize \[104\] into a product of perfect squares if possible. So, we will get substitute $4\times 26$ for \[104\] in the above step as:
$\Rightarrow diagonal=\sqrt{4\times 26}$
Since, $4$ is a perfect square of $2$. So, we will take out \[4\] outside of under root as:
$\Rightarrow diagonal=2\sqrt{26}$
Hence, the maximum length of pencil is $2\sqrt{26}$ that can be kept in a rectangular box of length $8$, breadth $6$ and height $2$.

Note: A cuboid is a three dimensional diagram and a cuboid contains twelve sides, twelve edges, eight vertices and 6 faces in its shape. The most important point is that the length, breadth and height of a cuboid does not seem to be equal. If the height, breadth and height would be equal, this shape would be a cube not cuboid.