Question 135051
Let x=the product that the radius needs to be decreased 
{{{A=pi*r^2}}} Solve for {{{r^2}}}

{{{r^2=A/pi}}}  take square root of each side

{{{r=+-sqrt(A/pi)}}} ------------------------eq1

Now, if we decrease the area by 75%, then we have 25% of the area left,so:

{{{(r-xr)=+-sqrt(0.25A/pi)}}} simplifying we have:

{{{(r-xr)=+-0.5sqrt(A/pi)}}}-----------------------------eq2

Now we can see from eq1 that {{{+-sqrt(A/pi)=r}}}. Substitute this into eq2

{{{r-xr=0.5r}}}  subtract r from each side

{{{r-r-xr=0.5r-r}}}  collect like terms
{{{-xr=-0.5r}}}  divide both sides by r
{{{x=0.5}}}------------this tells us that r has to be decreased by 50% to get a decrease of 75% in the Area

CK
{{{A1=pi*r^2}}}----------------eq1
{{{A2=pi*(0.5*r)^2}}} or
{{{A2=pi*0.25r^2}}}-------------eq2
{{{A2=0.25A1}}} because in eq2 {{{pi*r^2}}}=A1



Hope this helps---ptaylor