Bigger Cubes

If the side is s, then new side with 30% increase, $s'=1.3s$ Surface area of cube is $6 s^{2}$ New surface area= $6 s' \, ^{2} = 6 (1.3s)^{2} = 6 \times 1.69 s^{2}$ New/old ratio= 1.69, increase $= 0.69= 69 \%$

Actually, i always consider math as fun. May be, this is because i always try to give it a practical approach and solve the problems with as least amount of simple calculation as possible. And even in this case.., Assume a square of a round figured length, say 10 A 30% increase would in length would make it 130 on each side To calculate the area; Area before increase in length : 10 10=100 Area after increase in length : 13 13=169 Total in in area : 169-100=69

well, you just KNOW that the total area increases with the square of the linear dimensions, just like you KNOW the volume increases with the cube.

So if a linear dimension increased by 30%, that is like saying you multiplied the old edge length by 1.3

so therefor you have multiplied the AREA by (1.3)^2 = 1.69

so therefore the area has increased by 69%

This you just KNOW.

Now, work it out more politely

1.) say the edge length is E before we started 2.) therefore the area of one face of the cube is E E (E^2) 6.) and the cube has 6 faces, so the total surface area was 6 E^2

Now we increase E by a factor of 1.3

so the NEW area is 6 * (1.3 * E)^2

And the RATIO of the NEW area to the OLD area is then

increase = (6 * (1.3 E)^2 )/ (6 * E^2)

(1.3E) ^2 is the same as (1.3*E) * (1.3 *E)

we can rearrange that as 1.3 * 1.3 * E * E or 1.69 * E^2

giving us the ratio of

(1.69 * 6 * E^2) / (6 * E^2)

the (6*E^2) is in both the top and bottom and divides out (becomes 1), leaving

1.69

or a net increase of 69%, just like we 'new' when we started.

(1.3)^2-1=0.69. and 0.69*100=69%

Step 1 : Assume length of side to be 100 units. $SA = 6a^{2} = 60000 units^{2}$

Step 2 : Increasing length by 30% will bring the length of side to 130 units. $SA^{'} = 101400 units^{2}$

Step 3 : $SA^{'} - SA = 41400 units^{2}$

Step 4 : We have to figure out percentage increase and to do that we jut have to find out what percent of 60000 is equal to 41400 $x = \frac{41400}{60000} \times 100$ $Answer = \boxed{69}$

Surface Area= 6s² ...(A₁)

A₂ = 6(s + .30s)²

A₂ = 6(s + .30s)²

A₂/A₁ = 6(s + .30s)²/6s²

A₂/A₁ = 6s²(1.30)²/6s²

A₂/A₁ = (1.30)²

A₂/A₁ = 1.69

Therefore the area is increased by 69 %

6(10^2) = 100 6[(10*1.3)^2)=169 ((169/100)100%)-100=69%

Change in Surface Area $\Delta S=(\frac{6(\frac{130}{100}x)^2}{6x^2}-1)\times 100=69 \text{\%}$

Change in Volume $\Delta V=(\frac{(\frac{130}{100}x)^3}{x^3}-1)\times 100=119.70 \text{\%}$

if one side is 10X(let) of the smaller cube then surface area will be 600X^2 and after increasing the sides new surface area will be 1040X^2......hence ans will be 69%.