Question

Find the surface of a cube whose volume is 512m^3 .​

Answer 1

Answer:

VOLUME OF CUBE =L^3

3√L=3√512

L=8 m

TOTAL SURFACE AREA OF CUBE = 6L^2

6×8^2

6×64

384m^2= ans

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

Let assume that edge of the cube be a m.

As it is given that

$\rm \: Volume_{(Cube)} = 512 \: {m}^{3}$

We know, Volume of cube of edge a is given by

$\boxed{ \rm{ \:Volume_{(Cube)} = {a}^{3} \: }}$

So, using this result, we get

$\rm \: {a}^{3} = 512$

$\rm \: {a}^{3} = 8 \times 8 \times 8$

$\rm \: {a}^{3} = {8}^{3}$

$\color{green}\rm\implies \:a \: = \: 8 \: m$

Now,

$\rm \: Surface\:Area_{(Cube)} \: = \: {6a}^{2}$

$\rm \: = \: 6 \times {(8)}^{2}$

$\rm \: = \: 6 \times 64$

$\rm \: = \: 384 \: {m}^{2}$

Thus,

$\color{green}\rm\implies \:\boxed{ \rm{ \:Surface\:Area_{(Cube)} \: = \: 384 \: {m}^{2} \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$