Question

the maximum length of a pencil that kept be in a rectangular box of dimensions 12 cm into 9 cm into 8 cm is

Answer 1

Length of largest pencil that can be kept in a box

diagonal = $\sqrt{l^{2}+b^{2}+h^{2}}$

l = 12 , b = 9 , h = 8

length of largest pencil = $\sqrt{12^{2} +9^{2} + 8^{2}}$

= $\sqrt{289}$

= 17 cm

The maximum length of a pencil that kept be in a rectangular box of dimensions 12 cm into 9 cm into 8 cm is 17 cm

Answer 2

$\huge\underline\mathrm{GIVEN:-}$

• Length of rectangular box = 12 cm
• Breadth of rectangular box = 9 cm
• Height of rectangular box = 8 cm

$\huge\underline\mathrm{TO\: FIND:-}$

Find the length of the pencil = ?

$\huge\underline\mathrm{SOLUTION:-}$

We have given that, the dimensions of rectangular box is 12 cm $\times$9 cm $\times$8 cm

To find the length of the pencil, we use the formula:- $\bf\sqrt{{l}^{2} + {b}^{2} + {h}^{2}}$

Put the values in the formula

$\implies$$\bf\sqrt{{12}^{2} + {9}^{2} + {8}^{2}}$

$\implies$$\bf\sqrt{144 + 81 + 64}$

$\implies$$\bf\sqrt{289}$

$\implies$$\large\bf{17\:cm}$

Thus, the length of pencil kept in rectangular box is 17 cm