Question

If the volume of a cube is V m3, its surface area is S m2 and the length of a diagonal is
d metres, prove that 6√3 V = Sd.
​

Answer 1

Question : -

• If the volume of a cube is V m^3, its surface area is S m^2 and the length of a diagonal is d metres, prove that 6√3 V = Sd.

Answer :-

Given :-

• Volume of cube is V.
• Surface area of cube is S.
• Length of diagonal is d.

To Prove :-

• 6√3 V = Sd.

$\begin{gathered}\Large{\bold{\pink{\underline{CaLcUlAtIoN\::}}}} \end{gathered}$

$\begin{gathered}\bf\red{Let,} \end{gathered}$

$\begin{gathered}\longmapsto\:\:\bf{Edge \: of \: cube \:is \:a \: \:m}. \\ \end{gathered}</p><p>$

${{ \boxed{\large{\bold\red{Volume_{(Cube)}\: = a^3}}}}}$

${{ \boxed{\large{\bold\pink{Total \: Surface \: area_{(Cube)}\: = 6 a^2}}}}}$

${{ \boxed{\large{\bold\blue{Diagonal{(Cube)}\: = a\sqrt{3} }}}}}$

$\begin{gathered}\bf\pink{So,} \end{gathered}$

$\bf \:Volume, \: V = {a}^{3}$

$\bf \:Area, \: S = {6a}^{2}$

$\bf \:Diagonal, \: d = a \sqrt{3}$

$\begin{gathered}\Large\bf\pink{Consider} \end{gathered}$

$\bf \:6 \sqrt{3} V$

$\bf\implies \:6 \sqrt{3} \times {a}^{3}$

$\bf\implies \:6 \sqrt{3} \times {a}^{2} \times a$

$\bf\implies \:(6 \times {a}^{2} ) \times a \sqrt{3}$

$\bf\implies \:Sd$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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$\large{\boxed{\boxed{\bf{Explore \ more : - }}}}$

More info:

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²

Answer 2

Answer:

V=a^3

S=6a^2

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