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) (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.
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