Question

Maximum length of the rod that can be kept in a cuboid box of side 12 CM 9 cm and 8 cm

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• Length of cuboid, l = 12 cm

• Breadth of cuboid, b = 9 cm

• Height of cuboid, h = 8 cm

We have to find the maximum length of the rod that can be placed in the cuboid.

We know,

The maximum length of the rod that can be placed in the cuboid is equals to the diagonal of the cuboid.

Now, we know diagonal of a cuboid of length l, breadth b and height h is given by

$\boxed{ { \:Diagonal_{(Cuboid)} \: = \: \sqrt{ {l}^{2} + {b}^{2} + {h}^{2} } \: \: }}$

So, on substituting the values, we get

$\rm \: Diagonal_{(Cuboid)} = \sqrt{ {12}^{2} + {9}^{2} + {8}^{2} }$

$\rm \:Diagonal_{(Cuboid)} = \sqrt{ 144 + 81 + 64 }$

$\rm \: Diagonal_{(Cuboid)} = \sqrt{ 289}$

$\rm \: Diagonal_{(Cuboid)} = \sqrt{ {17}^{2} }$

$\rm\implies \:Diagonal_{(Cuboid)} = 17 \: cm$

$\bf\implies \:Maximum \: length \: of \: rod \: = \: 17 \: cm$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$