Question 282351
Let r equal the original radius.


The new radius is equal to 2r - 5


Let a equal the original area.


The new area is equal to a + 32*pi


The original area is given by the formula:


a = pi * r^2


The new area is given by the formula


a + 32 * pi = pi * (2r - 5)^2


Replace a with the equivalent expression of pi*r^2 and you get:


pi*r^2 + 32*pi = pi*(2r-5)^2


Simplify to get:


pi*r^2 + 32*pi = 4*pi*r^2 - 20*pi*r + 25*pi


Divide both sides of this equation by pi to get:


r^2 + 32 = 4r^2 - 20r + 25


Subtract r^2 + 32 from both sides of this equation to get:


3r^2 - 20r - 7 = 0


This factors out to be (3r+1) * (r-7) = 0


This makes r = -1/3 and r = 7


r can't be negative so the answer is r = 7.


that would be selection c.


Plug this value into the original equation and you get:


a = pi * r^2 becomes a = pi * 49 which becomes a = 49*pi.


The new radius is equal to 2r - 5 which makes the new radius equal to 14 - 5 which makes the new radius equal to 9.


a = pi * 9^2 becomes a = pi * 81 becomes a = 81*pi


81*pi - 49*pi = 32*pi satisfying the requirements of the problem that states that when the radius is increased from 7 to 9, the area is increased by 32*pi.


9 = 2*7 - 5 = 14 - 5 = 9 which is true, confirming the radius increase.


81*pi - 49*pi = 32*pi conforming the area increase.