Question

Each edge of a cube is increased by 40%,so the % increased in the surface area is

Answer 1

let m of edge be x
40/100×x+x=7/5a
increase in sa= 6×49/25xsq -6xsq
=(6xsq)×24/25
=% increase=(6xsq×24)/(6asq×25)×100
=96%

HOPE IT HELPS!

Answer 2

Answer:

The % increased in the surface area is 96%.

Step-by-step explanation:

Given : Each edge of a cube is increased by 40%.

To find : The % increased in the surface area ?

Solution :

Let the edge be 'a'.

The surface area of the cube is $SA=6a^2$

Each edge of a cube is increased by 40%.

i.e. increased edge is $s=a+40\%a$

$s=a+0.4a$

$s=1.4a$

The surface area of increased edge is

$SA_n=6(1.4a)^2$

$SA_n=1.96\times 6(a)^2$

The increased in surface area is given by,

$A=SA_n-SA$

$A=1.96\times 6(a)^2-6(a)^2$

$A=0.96\times 6(a)^2$

The percentage increase in surface area is given by,

$I\%=\frac{0.96\times 6(a)^2}{6a^2}\times 100$

$I\%=96\%$

Therefore, the % increased in the surface area is 96%.