SOLUTION: If the radius of a circle is increased so that it is 5 centimeters less than twice the original radius, the area is increased by 32π square centimeters. What is the original r

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Question 282351: If the radius of a circle is increased so that it is 5 centimeters less than twice the original radius, the area is increased by 32π square centimeters. What is the original radius?
a 3 b 1/3 c 7 d π/3 e 3π

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
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.