Question

ਚੰਦ ਦਾ ਵਿਆਸ ਧਰਤੀ ਦੇ ਵਿਆਸ ਦਾ ਇੱਕ ਚੌਥਾਈ ਹੈ।ਇਹਨਾਂ ਦੋਨਾਂ ਦੇ ਸਕੂਈ ਖੇਤਰਫਲਾਂ ਦਾ ਅਨੁਪਾਤ ਪਤਾ ਕਰੋ। The diameter of the moon is approximately one fourth of the diameter of the earth. find the ratio of thei surface areas.​

Answer 1

Answer:

Both the moon and earth are in the shape of spheres.

Surface area of a sphere of radius 'r' =4πr2

Let d1 be the diameter of the moon and d2 and be the diameter of the earth.

Let r1 be the radius of the moon and r2 be the radius of the earth.

Given d1=41d2

=>2r1=41×2r2

=>r1=41×r2

Now, ratio of their surface areas is:

S1:S2

=4πr12:4πr22

=r12:r22

=12:42

=1:16

Answer 2

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

Given that,

• The diameter of the moon is approximately one fourth of the diameter of the earth.

Let assume that

• Diameter of moon be D

So,

• Diameter of earth be 4D.

Thus, we have

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\:\begin{cases} &\sf{Radius_{moon} = \dfrac{D}{2} } \\ \\ &\sf{Radius_{earth} = 2D} \end{cases}\end{gathered}\end{gathered}$

We know,

$\\ \purple{\rm :\longmapsto\:\boxed{\tt{ SurfaceArea_{(sphere)} = {4\pi \: r}^{2}}}}$

where,

• r is surface area of the sphere

So,

$\rm :\longmapsto\:SurfaceArea_{(moon)} = 4\pi \: {\bigg(\dfrac{D}{2} \bigg) }^{2}$

$\bf\implies \:SurfaceArea_{(moon)} = \pi {D}^{2} - - - - - (1)$

Also,

$\rm :\longmapsto\:SurfaceArea_{(earth)} = 4\pi \: {(2D)}^{2}$

$\bf\implies \:SurfaceArea_{(earth)} = 16\pi {D}^{2} - - - - - (2)$

Thus,

$\rm :\longmapsto\:SurfaceArea_{(moon)} \: : \: SurfaceArea_{(earth)}$

$\rm \:  =  \: {\pi \: D}^{2} \: : \: {16\pi \: D}^{2}$

$\rm \:  =  \: 1\: : \:16$

Hence,

$\red{\boxed{\sf{ \:SurfaceArea_{(moon)} \: : \: SurfaceArea_{(earth)} = 1\: : \:16 \: }}}$

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Additional Information :-

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: CSA{(cuboid)} = 2(l + b) \times h}}$

$\boxed{ \sf{ \: CSA{(cube)} = 4 \times {(edge)}^{2} }}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$