Question

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

Answer 1

${\huge{\underline{\underline{\rm{Question:-}}}}}$

Q) A Cuboid of dimensions 60 cm × 54 cm × 30 cm . How many small cubes with side 6 cm can be placed in the given Cuboid .

${\huge{\underline{\underline{\rm{Answer\checkmark}}}}}$

★ Given :

• Dimensions of Cuboid = 60 cm × 54 cm × 30 cm
• Edge of cube = 6 cm

★ To Find :

• How many such cubes can be placed in the Cuboid .

★ Solution :

We know ,

• Length (Cuboid) = 60 cm
• Breadth (Cuboid) = 54 cm
• Height (Cuboid) = 30 cm

So ,

Using the Formula ↓

$\underline{ \sf \: volume \: of \: a \: cuboid = length \times breadth \times height}$

Here ,

$\underline{ \boxed{ \rm \: volume_{cuboid} = 60 \times 54 \times 30 \: {cm}^{3} }}$

And ,

Using the Formula :

$\underline{ \sf \: volume \: of \: a \: cube = {edge}^{3} }$

Here ,

$\underline{ \boxed{ \rm \: volume _{cube} = {6}^{3} = 6 \times 6 \times 6 \: {cm}^{3} }}$

Now , We know that :

$\rm \: the \: no. \: of \: cubes \: that \: can \: be \: placed = \frac{volume _{cuboid} }{volume _{cube} }$

$\mapsto \rm \: no. \: of \: cubes = \frac{ \cancel{60} \times \cancel{54} \times \cancel{30} \: \cancel {cm}^{3} }{ \cancel6 \times \cancel6 \times \cancel6 \: \cancel {cm}^{3} }$

$\mapsto \rm \: no. \: of \: cubes = 10 \times 9 \times 5$

$\large \mapsto \pink{ \boxed {\purple{{ \sf \: no. \: of \: cubes = 450}}}}$

So ,

450 such cubes can be placed in the Cuboid .

_________________________

★ Some Important Formulas :

• Volume of Cuboid = Length×Breadth×Height

• Volume of Cube = Side³

• TSA of Cuboid = 2(LB + BH + LH)

• TSA of Cube = 6(side)²

• CSA of Cuboid = 2(L+B)×H

• CSA of Cube = 4(side)²

Answer 2

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

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

$\sf \mapsto{Dimension \: of \: cuboid \: is \: 60cm \times 54cm \times 30cm.}$

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

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

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

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

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

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

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

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

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

${ \implies{\sf{Volume \: of \: cuboid=60cm×54cm×30cm }}}$

${ \implies{\sf{Volume \: of \: cuboid=97,200 {cm}^{3} }}}$

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

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

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

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

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

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \frac{97,200 {cm}^{3} }{216 {cm}^{3} } }}}$

${ \implies{\sf{No. \: of \: small \: cubes \: placed \: in \: large \: cuboid= \cancel{\frac{97,200 {cm}^{3} }{216 {cm}^{3} }} = 450cubes}}}$

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

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

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

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

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

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