Question

Two cubes have their volumes in the ratio 1:27 . What is the ratio of their surface areas..

Answer 1

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QUESTION -

Two cubes have their volumes in the ratio 1:27 . What is the ratio of their surface areas..

SOLUTION -

Let the edge of the two cubes be and b respectively.

We know that :

Volume of a cube = { Side } ^ 3

So,

Volume of 1st cube. = { a } ^ 3.

Volume of 2nd cube =. { b } ^ 3

Now we have the following information.....

Two cubes have their volumes in the ratio 1:27

$=>{ ( \dfrac{ a}{b} ) } ^ 3 = \dfrac{1}{27} \\ \\ => \dfrac{ a}{b} = \dfrac{1}{3}$

Now,

The surface area of a cube is 6 { side } ^ 2.

Surface area of Cube 1 = 6 a ^ 2.

Surface area of Cube 2 = 6 b ^ 2

$\dfrac{ 6 { a } ^ 2 } { 6 { b } ^ 2 } = \dfrac{ { a } ^ 2 }{ { b } ^ 2 } = \dfrac{ 1}{9} .............. [ A ]$

Answer 2

Given :-

Two cubes have their volumes in the ratio 1:27.

To find :-

The ratio of their surface areas.

Solution :-

We know,

Volume of a cube = (side)³

A/q,

$\displaystyle{\sf{\frac{a^3}{A^3}=\frac{1}{27} }}$

Taking (³) to right side :

$\displaystyle{\sf{\implies \frac{a}{A}=\frac{1}{3} }}$

So, the ratio of the edges of the cubes is 1 : 3.

Now, surface area of a cube = 6a².

Ratio of the surface areas :

$\displaystyle{\sf{\frac{6a^2}{6A^2} }}\\\\\displaystyle{\sf{=\frac{a^2}{A^2} }}\\\\\displaystyle{\sf{=(\frac{a}{A} )^2}}\\\\\displaystyle{\sf{=(\frac{1}{3})^2 }}\\\\\huge\bold{\boxed{\displaystyle{\sf{=\frac{1}{9} }}}}$

∴ So, the ratio of their surface areas is 1 : 9.