Question

2 cubes have volume 27:64. find the ratio of their surface areas

Answer 1

Let the side of 1st cube be a
& the 2nd cube be b

$\frac{vol \: of \: 1st \: cube}{vol \: of \: 2nd \: cube} = \frac{27}{64}$


a³/b³=27/64

$\frac{a}{b } = \frac{3}{4}$


T.S.A of 1st /T.S.A of 2nd=6a²/6b²

=a²/b²

=(3)²/(4)²

=9/16

T.S.A of 1st:T.S.A of 2nd=9:16

Answer 2

Answer:

The ratio of their surface areas are 9:16

Step-by-step explanation:  

Given that 2 cubes have volume 27:64.

we have to find the ratio of their surface areas

Let the side of 1st cube be a  

and the side of 2nd cube be b

$\text{volume of cube of side a=}a^3$

$\frac{\text{Volume of 1st cube}}{\text{volume of 2nd cube}}= \frac{27}{64}$

$\frac{a^3}{b^3}=\frac{27}{64}$

$\frac{a}{b}=\frac{3}{4}$

$\frac{\text{T.S.A of 1st cube}}{\text{T.S.A of 2nd cube}}=\frac{6a^2}{6b^2}=\frac{a^2}{b^2}$

$=\frac{3^2}{4^2}=\frac{9}{16}$

T.S.A of 1st : T.S.A of 2nd=9:16