Question

The radius of a sphere increased by $50$ percent. By how many percent did the surface area of the sphere increase?

Answer 1

Let the Initial surface area $S=4\pi R^2$

$R^{\prime }=R+\frac{R\times 50}{100}=\frac{3R}{2}$

New surface area $S^{\prime }=4\pi R^{\prime 2}$

$\frac{S^{\prime }}{S}=\frac{4\pi R^{\prime 2}}{4\pi R^2}$

$=\frac{R^{\prime 2}}{R^2}=\frac{3\times 3}{2\times 2}$

$=\frac{9}{4}$

$⟹S^{\prime }=2.25S$

$⟹\Delta S=S^{\prime }-S=1.25S$

Hence percentage increase $=\frac{\Delta S}{S}\times 100=\frac{1.25S}{S}\times 100$

$=125\%$