Question

4. A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with side
6 cm can be placed in the given cuboid?​

Answer 1

Answer:

• 450 small cubes with side 6 cm can be placed in the given cuboid.

Step-by-step explanation:

Given that:

• A cuboid is of dimensions 60 cm x 54 cm x 30 cm.

To Find:

• How many small cubes with side 6 cm can be placed in the given cuboid?

Formula used:

• Volume of a cuboid = (Length x Breadth x Height)
• Volume of a cube = (side x side x side)

Finding the volume of a cuboid:

⟶ Volume of a cuboid = (60 cm x 54 cm x 30 cm)

⟶ Volume of a cuboid = 97200 cm³

Finding the volume of a cube of side 6 cm:

⟶ Volume of a cube = (6 cm x 6 cm x 6 cm)

⟶ Volume of a cube = 216 cm³

Finding the number of cubes placed in the cuboid:

⟶ Number of cubes = (Volume of cuboid)/(Volume of cube)

⟶ Number of cubes = (97200 cm³)/(216 cm³)

⟶ Number of cubes = 450

Answer 2

Answer:

$\large\underline{\sf{\pmb{Given}}}$

• ➠ A cuboid is of dimensions 60 cm x 54 cm x 30 cm.

$\begin{gathered} \: \: \end{gathered}$

$\large\underline{\sf {\pmb{To \: Find}}}$

• ➠ How many small cubes with side 6 cm can be placed in the given cuboid?

$\begin{gathered} \: \: \end{gathered}$

$\large\underline{\sf{ \pmb{Using \: Formula }}}$

$\bigstar \: \underline{ \boxed{\sf {Volume \: of \: a \: cuboid = (Length \times Breadth \times Height) }}}$

$\bigstar \: \underline { \boxed{\sf{{Volume \: of \: a \: cube = ( {a}^{3} ) }}}}$

$\bigstar \: \underline{ \boxed{\sf{{Required \: number \: of \: Cube = \dfrac{Volume \: of \: Cuboid}{volume \: of \: the \: small \: cube} }}}}$

Where

• $\leadsto \sf{a = Side }$

$\begin{gathered} \: \: \end{gathered}$

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

Firstly finding the volume of a cuboid

${ \implies\sf{Volume \: of \: a \: cuboid = (Length \times Breadth \times Height) }}$

• Substituting the values

${\implies \sf{ Volume_{(cuboid) }= (60 cm \times 54 cm \times 30 cm) }}$

${\implies \sf { Volume_{(cuboid) }= 97200 \: {cm}^{3} }}$

$\: \: \: \star \: \underline {\boxed{\sf \purple{{volume \: of \: a \: cuboid = 97200 \: {cm}^{3} }}}} \: \star$

$\begin{gathered} \: \: \end{gathered}$

Now Finding the volume of a cube

$\implies{\sf{{Volume \: of \: a \: cube = ( {a}^{3} ) }}}$

• Substituting the values

${ \implies \sf{Volume_{(cube)} = (6 cm \times 6 cm \times 6 cm) }}$

${ \implies \sf{Volume_{(cube)} = 216 \: {cm}^{3} }}$

$\: \: \: \star \: \underline {\boxed{\sf \purple{{Volume \: of \: a \: cube = 216 \: {cm}^{3} }}}} \: \star$

$\begin{gathered} \: \: \end{gathered}$

Finding the number of cubes placed in the cuboid

${\implies\sf{{Required \: number \: of \: Cube = \dfrac{Volume \: of \: Cuboid}{volume \: of \: the \: small \: cube} }}}$

• Substituting the values

$\implies \sf Number \: of \: cubes = \dfrac{97200}{216}$

$\implies \sf Number \: of \: cubes = \cancel\dfrac{97200}{216}$

$\implies\sf{Number \: of \: cubes = 450}$

$\: \: \: \large \star \: \underline {\boxed{\sf \purple{Number \: of \: cubes = {450}}}} \: \star$

$\begin{gathered} \: \: \end{gathered}$

$\large\underline{ \sf{\pmb{ Therefore}}}$

• ➠ Total 450 small cubes with side 6 cm can be placed in the given 60 cm x 54 cm x 30 cm cuboid.

$\begin{gathered} \: \: \end{gathered}$

$\large \underline{\sf{ \pmb{Additional \: Information }}}$

$\begin{gathered}\begin{gathered}\bigstar \: \bf\underline{More \: Useful \: Formulae } \: \bigstar  \\ \begin{gathered}{\boxed{\begin{array} {cccc}{\sf{{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}} \\  \\{\sf{{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}  \\  \\{\sf{{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}} \\  \\ {\sf{{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}} \\  \\ {\sf{{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\  \\ {\sf{{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\  \\{\sf{{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\  \\ {\sf{{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\  \\ {\sf{{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\  \\ {\sf{{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\  \\ {\sf{{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\  \\{\sf{{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}\end{array}}}\end{gathered}\end{gathered}\end{gathered}$