Question

3. Find the surface area of a cube whose edge is :
(i) 4 m
​

Answer 1

Given : Edge of a cube is 4 m .

Need To Find : Lateral Surface Area of Cube.

$\underline {\frak{ As, \:We \:know\:that,\:}}$

$\qquad \large {\boxed { \pink {\sf{\bigstar Lateral\:Surface \:Area \:_{(Cube)} = 4 \times (a)^{2} \:sq.units }}}}$

Where ,

• a is the edge of a cube in m .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad \qquad \qquad:\implies \sf{ L.S.A _{(Cube)} = 4 \times 4^{2}}$

$\qquad \qquad \qquad:\implies \sf{ L.S.A _{(Cube)} = 4 \times 16}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  L.S.A _{(Cube)} = 64\: m^{2}}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Lateral \:Surface \:Area\:of\:Cubr \:is\:\bf{64\: m^{2}}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Answer 2

Explanation:

Given : Edge of a cube is 4 m .

Need To Find : Lateral Surface Area of Cube.

\begin{gathered}\underline {\frak{ As, \:We \:know\:that,\:}}\\\end{gathered}

As,Weknowthat,

\begin{gathered}\qquad \large {\boxed { \pink {\sf{\bigstar Lateral\:Surface \:Area \:_{(Cube)} = 4 \times (a)^{2} \:sq.units }}}}\\\end{gathered}

★LateralSurfaceArea

(Cube)

=4×(a)

2

sq.units

Where ,

a is the edge of a cube in m .

⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\\end{gathered}

⋆NowBySubstitutingtheGivenValues:

\begin{gathered}\qquad \qquad \qquad:\implies \sf{ L.S.A _{(Cube)} = 4 \times 4^{2}}\\\end{gathered}

:⟹L.S.A

(Cube)

=4×4

2

\begin{gathered}\qquad \qquad \qquad:\implies \sf{ L.S.A _{(Cube)} = 4 \times 16}\\\end{gathered}

:⟹L.S.A

(Cube)

=4×16

⠀⠀⠀⠀⠀\begin{gathered}\underline {\boxed{\pink{ \mathrm { L.S.A _{(Cube)} = 64\: m^{2}}}}}\:\bf{\bigstar}\\\end{gathered}

L.S.A

(Cube)

=64m

2

★

Therefore,

⠀⠀⠀⠀⠀\begin{gathered}\therefore {\underline{ \mathrm { Lateral \:Surface \:Area\:of\:Cubr \:is\:\bf{64\: m^{2}}}}}\\\end{gathered}

∴

LateralSurfaceAreaofCubris64m

2

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

\begin{gathered}\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\\end{gathered}

∣

⋆MoreToknow:

∣

\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}

†

FormulasofAreas:−

⋆Square=(side)

2

⋆Rectangle=Length×Breadth

⋆Triangle=

2

1

×Breadth×Height

⋆Scalene△=

s(s−a)(s−b)(s−c)

⋆Rhombus=

2

1

×d

1

×d

2

⋆Rhombus=

2

1

p

4a

2

−p

2

⋆Parallelogram=Breadth×Height

⋆Trapezium=

2

1

(a+b)×Height

⋆EquilateralTriangle=

4

3

(side)

2