Question

A cuboid is of dimensions 60cm × 40cm × 20cm.How many small cubes with side 4cm can be placed in the given cuboid? ​

Answer 1

Answer:

750 cubes

Step-by-step explanation:

So here ,

Volume of the cuboid- 60×40×20

Volume of one small cube - a cube

4×4×4

Number of cubes =Volume of cuboid/Volume of cube

So ,

60×40×20/4×4×4

= 48000/64

=750 cubes

Hence 750 cubes can be placed in the given cuboid .

Hope it helped !!!

Answer 2

$\Huge \underline \red{{\fbox{Answer}}}$

$\large \underline{ \underline {\red {\frak{Given::}}}}$

$\sf \mapsto{Dimension \: of \: cuboid \: is \: 60cm \times 40cm \times 20cm.}$

$\sf \mapsto{Side \: of \: small \: cube \: is \: 4cm.}$

$\large \underline{ \underline {\red {\frak{To \: find::}}}}$

$\sf \leadsto{No. \: of \: small \: cube \: placed \: in \: large \: cuboid.}$

$\large \underline{ \underline {\red {\frak{Formula \: required::}}}}$

${\pink{\star}} \underline{\boxed{\bf{Volume \: of \: cuboid=lbh }}} {\pink{ \star}}$

${\pink{\star}} \underline{\boxed{\bf{Volume \: of \: cube={l}^3 }}} {\pink{ \star}}$

${\pink{\star}} \underline{\boxed{\bf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \frac{Volume \: of \: large \: cuboid}{ Volume \: of \: small \: cube} }}} {\pink{ \star}}$

$\large \underline{ \underline {\red {\frak{According \: to \: Question::}}}}$

${ \implies{\sf{Volume \: of \: cuboid=lbh }}}$

${ \implies{\sf{Volume \: of \: cuboid=60cm×40cm×20cm }}}$

${ \implies{\sf{Volume \: of \: cuboid=48,000 {cm}^{3} }}}$

$\implies{\sf{Volume \: of \: cube={l}^{3} }}$

$\implies{\sf{Volume \: of \: cube={(4)}^{3} }}$

$\implies{\sf{Volume \: of \: cube=4cm \times4cm\times4cm}}$

$\implies{\sf{Volume \: of \: cube=64{cm}^{3} }}$

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \frac{Volume \: of \: large \: cuboid}{Volume \: of \: small \: cube} }}}$

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \frac{48,000 {cm}^{3} }{64{cm}^{3} } }}}$

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \cancel{\frac{48,000 {cm}^{3} }{64 {cm}^{3} }} = 750cubes}}}$

$\large \underline{ \underline {\red {\frak{Know \: more::}}}}$

$\boxed{ \begin{array}{|l|l| } \hline \sf T.S.A. \: of \: cuboid& \sf2(lb + bh + hl) \\ \hline \sf L. S. A. \: of \: cuboid& \sf2h(l + b) \\ \hline \sf Diagonal \: of \: cuboid& \sf\sqrt{ {l}^{2} + {b}^{2} + {h}^{2}} \\ \hline \sf T.S.A. \: of \: cube& \sf 6( {l}^{2} ) \\ \hline \sf L. S. A. \: of \: cube& \sf4( {l}^{2} ) \\ \hline\sf Diagonal \: of \: cube& \sf\sqrt{3}l \\ \end{array}}$

$\large \underline{ \underline {\red {\frak{Note::}}}}$

$\sf{Here \: we \: use \: to \: {\bf T. S. A.}, { \bf L. S. A.} ,{ \bf l}, { \bf b} \: and \: { \bf h}}$

$\sf{to \: represent \: {\bf total \: surface \: area}, {\bf lateral \: surface \: area}, {\bf length},}$

$\sf{ {\bf breadth} \: and \: {\bf{height}} \: respectively. }$