Question

Find the volume of a cuboid whose
length
= 3.5 m, b = 2.6 m and h = 90 cm.

​

Answer 1

The volume of a cuboid of length l, breadth b, and height h =l×b×h

So, the volume of the cuboid with length =3.5m, breadth =2.6m and height =90cm or 0.9m

=3.5×2.6×0.9=8.19m3

Total surface area of a cuboid of length l, breadth b and height h =2(l×b+b×h+l×h)

So, the total surface area of the cuboid with length =3.5m, breadth =2.6mand height =90cm or 0.9m

=2(3.5×2.6+2.6×0.9+3.5×0.9)=2(9.1+2.34+3.15)=29.18m2 

Answer 2

$\Huge \fbox{Answer}$

$\Large \underline{\underline{\frak {\color{red}Given::}}}$

$\mapsto\sf{Length \: of \: cuboid = 3.5m}$

$\mapsto\sf{Breadth \: of \: cuboid = 2.6m}$

$\mapsto\sf{Height \: of \: cuboid = 90cm}$

$\Large \underline{\underline{\frak {\color{blue}To \: find::}}}$

$\leadsto\sf{Volume \: of \: cuboid.}$

$\Large \underline{\underline{\frak {\color{indigo}Formula \: used::}}}$

$\maltese\boxed {\rm{Volume \: of \: cuboid = length(l) \times breadth(b) \times height(h)}} \maltese$

$\Large \underline{\underline{\frak {\color{green}Concept \: required::}}}$

$\rightarrow \sf{1cm = \frac{1}{100} m}$

$\rightarrow \sf{90cm = \frac{90}{100}m = 0.9m }$

$\Large \underline{\underline{\frak {\color{peru}According \: to \: Question::}}}$

${ : :\implies{\sf{Volume \: of \: cuboid = length(l) \times breadth(b) \times height(h)}} }$

${ : :\implies{\sf{Volume \: of \: cuboid = 3.5m \times 2.6m \times 0.9m}} }$

${ : :\implies{\sf{Volume \: of \: cuboid = 8.19 {m}^{3} }} }$

$\Large{\underline{\frak {\color{gold}Hence, }}}$

$\star{\boxed{\sf{Volume \: of \: cuboid = 8.19 {m}^{3} }}} \star$

$\Large \underline{\underline{\frak {\color{si}Know \: more::}}}$

${ \boxed{ \begin{array}{ |l|l } \hline \sf Total \: surface \: area \: of \: cubiod& \sf 2(length \times breadth + breadth \times height + height \times length) \\ \hline \sf Lateral \: surface \: area& \sf 2 \times height(length + breadth) \\ \hline \sf Diagonal \: of \: cuboid& \sf \sqrt{ {length}^{2} + {breadth}^{2} + {height}^{2} } \\ \end{array}}}$

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