Question

Each edge of a cube is increased by 50%. Find the percentage increase in the

surface area of the cube.​

Answer 1

Let the initial side length be x. Then the surface area of cube is

A_1=6x^2A

1

=6x

2

If each side of a cube is increased by 50%, then the new side length is

x+\frac{50}{100}x=x+0.5x=1.5xx+

100

50

x=x+0.5x=1.5x

The surface area of new cube is

A_2=6(1.5x)^2A

2

=6(1.5x)

2

A_2=6(1.5)^2x^2A

2

=6(1.5)

2

x

2

A_2=13.5x^2A

2

=13.5x

2

The percentage increase in its surface area is

P=\frac{A_2-A_1}{A_1}\times 100P=

A

1

A

2

−A

1

×100

P=\frac{13.5x^2-6x^2}{6x^2}\times 100P=

6x

2

13.5x

2

−6x

2

×100

P=\frac{7.5}{6}\times 100=125%P=

6

7.5

×100=125

Therefore the surface area increased by 125%.

Answer 2

let edge of cube be a

surface area = 6a^2 = A let

new edge = 1.5a (50% increased)

new surface area

= 6*(1.5a)^2

= 2.25*(6a^2)

= 2.25A

increase in area = 2.25A - A

= 1.15. A

percent increase = 1.15A*100/A

= 115%