Question

Each edge of a cube is increased by 50% .Find the percentage increase in the surface area. Explain in detail. plzzzzz

Answer 1

Increase by 50%=a+a/2=3a/2
Total surface area of original cube=6a^2
TSA of new cube =6(3a/2)^2=6(9a^2/4)
=13.5a^2
Increase in area=13.5a^2-6a^2=7.5a^2
7.5a^2 Increase %=7.5a^2/6a^2×100
=125%

Answer 2

Answer:

$\bold{108\frac{1}{3}}$ %

Step-By-Step Explanation:

In such questions, the first  step of all individuals should be assuming the side to be a variable.

Let the measure of each edge of the cube be ' x ' cm.

Total surface area of the cube = 6 ( x ) ^ 2

Total surface area of the cube = 6 * x ^ 2

Total surface area of the cube = 6x ^ 2 cm²

Now when each edge is increased by 50 %,

Measure of the edge of new cube = x + 50 % of x

Measure of the edge of new cube = x + 50x / 100

Measure of the edge of new cube = x + x / 2

Measure of the edge of new cube = ( 3x / 2 ) cm

Total surface area of the new cube = 6 * ( 3x / 2 ) ^ 2

Total surface area of the new cube = 6 * ( 9x^2 / 4 )

Total surface area of the new cube = 54x ^ 2 / 4

Total surface area of the new cube = 12.5 x ^ 2 cm²

Increase in the total surface area = 12.5 x ^ 2 - 6 x ^ 2

Increase in the total surface area = 6.5 x ^ 2 cm²

Now,

Percentage increase in the total surface area = ( Increase in total surface area * 100 ) / Initial total surface area of the cube

Percentage increase in the total surface area = ( 6.5 x ^ 2 * 100 ) / 6 x ^ 2

Percentage increase in the total surface area = 650 x ^ 2 / 6 x ^ 2

Percentage increase in the total surface area = 650 / 6

Percentage increase in the total surface area = 325 / 3

Percentage increase in the total surface area = $\bold{108\frac{1}{3}}$ %