Question

Each edge of a cube is increase by 25% find the percentage increase in the t.s.a

Answer 1

Question:

Each edge of a cube is increase by 25% find the percentage increase in the Total surface Area (T.S.A.) ?

Answer:

56.25 %

Note:

If the edge length of a cube is a units , then ;

• Volume (V) = a^3 cubic units

• Lateral surface area (L.S.A.) = 4•a^2 sq. units

• Total surface area (T.S.A.) = 6•a^2 sq. units

• Percentage increase = ( Q2/Q1 - 1)•100 %

where , Q2 denotes the final quantity

and Q1 denotes the initial quantity.

Solution:

Let the edge length of the cube be 100 units.

Thus ,

=> Initial edge (a1) = 100 units

Also;

=> Initial T.S.A (s1) = 6•(a1)^2 sq. units

=> s1 = 6•(100)^2 sq. units

When ,the edge of cube is increased by 25% then ;

=> Final edge (a2) = (100 + 25% of 100) units

=> a2 = [100 + (25/100)•100] units

=> a2 = (100 + 25) units

=> a2 = 125 units

Also,

=> Final T.S.A.(s2) = 6•(a2)^2 sq. units

=> s2 = 6•(125)^2 sq. units

Now,

The percentage increase in the Total surface area of the cube will be given as;

= [ s2/s1 - 1 ]•100 %

= [ {6•(125)^2}/{6•(100)^2} - 1 ]•100 %

= [ (125)^2/(100)^2 - 1 ]•100 %

= [ (125/100)^2 - 1 ]•100 %

= [ (5/4)^2 - 1 ]•100 %

= [ 25/16 - 1 ]•100 %

= [ (25 - 16)/16]•100 %

= [ 9/16 ]•100 %

= 900/16 %

= 56.25 %

Hence,

The percentage increase in the total surface area of the cube will be 56.25 % .

Answer 2

$\huge\mathfrak\blue{Answer:-}$

Given:

Each edge of a cube is increase by 25%

To Find:

The percentage increase in the total surface area.

Solution:

Let the sides of the cube be s.

The surface area is :

6 × s² = 6s²

When the length is increased by 25%.

The new length is :

1.25s

The surface area :

6 × 1.25s × 1.25s = 9.375s²

The increase in Surface Area = 9.375s² - 6s²

= 3.375s²

Percentage increase = 3.375s²/6s² × 100 = 56.25%

= 56.25%

Hence, the percentage increase in the total surface area is 56.25%.