If the radius of a sphere is increased by 40%, then find the percentage increase in its surface area.

To find the percentage increase in the surface area of a sphere when its radius is increased, we can follow these steps:

Let the original radius of the sphere be _r.
The formula for the surface area (_A) of a sphere is given by:
$A=4\pi r^2$

If the radius is increased by 40%, the new radius (_r') becomes:
$r^{\text{'}}=r+0.40r=1.4r$

Now, let us calculate the new surface area (_A') of the sphere using the new radius:
$A^{\text{'}}=4\pi {\left(1.4r\right)}^2$
$A^{\text{'}}=4\pi \left(1.96r^2\right)$
$A^{\text{'}}=1.96\left(4\pi r^2\right)=1.96A$

The increase in the surface area is the difference between the new surface area and the original surface area:
$\text{Increase in Area}=A^{\text{'}}-A=1.96A-A=0.96A$

To express this increase as a percentage of the original surface area, we calculate:
$\text{Percentage Increase}=\left(\frac{0.96A}{A}\right)\times 100=96\%$

Therefore, the percentage increase in the surface area of the sphere is 96%.