Question

please solve this question​

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

Therefore the surface area increased by 125%.

Answer 2

HRYA HERE'S YOUR ANSWER

let the length of one side be 2cm

then surface area=6×2×2=24cm^3

after 50% is increased ,the side becomes 3cm

then,surface area =6×3×3=54cn^2

PERCENTAGE increased =54-24/24×100

=30/24×100

=125%

HOPE IT HELPS