Question

A cuboid of dimensions 75cm×54cm×60cm.How many small cubes with side 6 cm can be placed in the given cuboid

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Answer 1

Step-by-step explanation:

area of cuboid = 2(lb+bh+hl)

= 2(75×54 + 54x60 + 75x60)

= 2 (4050 + 3240 + 4500)

= 2( 11,790)

= 23,580

Area of the cube = 6( side) (side)

= 6(6)(6)

=216 sq. cm.

now, the number of cubes = area of cuboid / area of cube

= 23,580 / 216

= 109 cubes can be made.

Answer 2

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

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

$\sf \mapsto{Dimension \: of \: cuboid \: is \: 75cm \times 54cm \times 60cm.}$

$\sf \mapsto{Side \: of \: small \: cube \: is \: 6cm.}$

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

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

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

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

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

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

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

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

${ \implies{\sf{Volume \: of \: cuboid=75cm×54cm×60cm }}}$

${ \implies{\sf{Volume \: of \: cuboid=2,43,000 {cm}^{3} }}}$

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

$\implies{\sf{Volume \: of \: cube={(6)}^{3} }}$

$\implies{\sf{Volume \: of \: cube=6cm \times6cm\times6cm}}$

$\implies{\sf{Volume \: of \: cube=216{cm}^{3} }}$

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

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \frac{2,43,000 {cm}^{3} }{216 {cm}^{3} } }}}$

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \cancel{\frac{2,43,000 {cm}^{3} }{216 {cm}^{3} }} = 1,125cubes}}}$

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

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

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

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

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

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