Question

Volume of two cubes are in the ratio 64 : 27. The ratio of their surface areas is :​

Answer 1

Answer:HERE IS UR ANSWER

Step-by-step explanation::​Volume of two cubes V1:V2 = 64:27

That is a13 : a23 = 43 : 33

⇒ a1 : a2 = 4 : 3

Therefore,  a1 / a2 = 4 / 3

Ratio of surface areas is  6a12 : 6a22  

=  a12 : a22  

= (a1 / a2)2  

= (4 / 3)2

= 16 : 9

HOPE IT IS HELPFULL TO U

Answer 2

${\huge{\underline{\underline{\sf{\maltese Question}}}}}$

Volume of two cubes are in the ratio 64 : 27 . The ratio of their Surface areas is ?

${\huge{\underline{\underline{\sf{\maltese Answer\checkmark}}}}}$

Let there be two cubes

• Cube 1 &
• Cube 2

# Given :

• $\rm V_{1}$ : $\rm V_{2}$=64:27

# To Find :

• $\rm TSA_{1} : TSA_{2}$ = ?

# Solution :

Since ratios of volumes are given ,

$\rm \: \frac{v_{1} }{v _{2}} = \frac{64}{27}$

We know that :

$\boxed{\rm \: volume \: of \: cube = {(side)}^{3} }$

So ,

$\rm \frac{ {side_{1} }^{3} }{{side_{2}}^{3} } = \frac{64}{27}$

$\mapsto \rm \: {( \frac{side_{1}}{side_{2}} )}^{3} = \frac{4 \times 4 \times 4}{3 \times 3 \times 3}$

$\mapsto \rm \: \frac{side _{1} }{side_{2} } = \sqrt[3]{( \frac{4 \times 4 \times 4}{3 \times 3 \times 3} )}$

$\mapsto \boxed{\rm \: \frac{side_{1}}{side_{2} } = \frac{4}{3} }....(i)$

Now ,

We have to find the ratio of TSA s

$\rm \: \frac{tsa _{1} }{tsa _{2}} = \frac{ \cancel6 \times {(side _{1})}^{2} }{ \cancel6 \times {(side_{2}) }^{2} }$

$\implies \rm \: \frac{tsa_{1}}{tsa_{2} } = \frac{ {(side_{1}) }^{2} }{ {(side_{2}) }^{2} }$

$\implies \rm \: \frac{tsa_{1}}{tsa_{2}} = {( \frac{side_{1} }{side_{2} }) }^{2}$

now , from equation (i)

$\implies \rm \: \frac{tsa _{1} }{tsa_{2} } = {( \frac{4}{5} )}^{2}$

$\implies \boxed{ \rm \: \frac{tsa_{1} }{tsa _{2}} = \frac{16}{9} }$

So ,

The ratio of their Surface areas would be 16 : 9 .