Question

If the surface area of a cube is increased by 1% what is the percentage increase in its volume

Answer 1

please mark brainliest

Answer 2

The percentage increase in the volume is 1.505%.

Formula:

The surface area of the cube is $6a^{2}$.

the volume of the cube is $a^{3}$.

Given:

The surface area of a cube is increased by 1%.

Explanation:

Let the surface area of the cube is $6a^{2}$.

The surface area of the new cube is $6b^{2}$.

The new surface area will be = 1% of old surface area + old surface area

New area,

$6b^{2}$ = 1.01× $6a^{2}$

$b=\sqrt{1.01}$ × $a$

Now,  the volume of first cube is  $V_{1} =a^{3}$.

therefore the new volume is  $V_{2} =b^{3\\}$.

$b^{3} = \sqrt{1.01}^{3} a^{3}$

Percentage increase in volume is = $\frac{V_{2}-V_{1} }{V_{2} }$×$100$

Putting the values of  $V_{1} =a^{3}$ and  $V_{2} =b^{3\\}$ in above and substitute the value of $b=\sqrt{1.01}$ × $a$  in them,

Percentage increase in volume comes out to be 1.505%.

#SPJ2