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Two circles with radii r and r + 2 have areas that differ by 8π. What

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Two circles with radii r and r + 2 have areas that differ by 8π. What [#permalink]

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New post 27 Nov 2017, 22:33
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A
B
C
D
E

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Question Stats:

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Re: Two circles with radii r and r + 2 have areas that differ by 8π. What [#permalink]

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New post 27 Nov 2017, 23:10
C

Area of larger circle = pi.(r+2)^2
Area of smaller circle =pi.r^2

Difference is pi.((r+2)^2-r^2)=8.pi

Solving we get r=1. Hence radius of larger circle =3


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Re: Two circles with radii r and r + 2 have areas that differ by 8π. What [#permalink]

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New post 28 Nov 2017, 14:08
Bunuel wrote:
Two circles with radii r and r + 2 have areas that differ by 8π. What is the radius of the larger circle?

(A) 1
(B) 2
(C) 3
(D) 8
(E) 9

Two circles with radii r and (r + 2) have areas that differ by 8π:

\(π(r+2)^2 - (πr^2) = 8π\)

\(π(r^2 + 4r + 4) - πr^2 = 8π\)

\(πr^2 + 4πr + 4π - πr^2 = 8π\)
\(4πr + 4π = 8π\)
\(4πr = 4π\)
\(r = 1\)


Radius of larger circle = \((r + 2) = 3\)

Answer C

Check: Circle 1 area = \(π(1)^2 = π\). Circle 2 area = \(π(3)^2 = 9π\). Difference:\((9π-1π)=8π\)

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Re: Two circles with radii r and r + 2 have areas that differ by 8π. What [#permalink]

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New post 29 Nov 2017, 17:18
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Expert's post
Bunuel wrote:
Two circles with radii r and r + 2 have areas that differ by 8π. What is the radius of the larger circle?

(A) 1
(B) 2
(C) 3
(D) 8
(E) 9


The area of the smaller circle is πr^2.

The area of the larger circle is π(r + 2)^2 = π(r^2 + 4r + 4).

Since the areas differ by 8π, we have:

π(r^2 + 4r + 4) - πr^2 = 8π

r^2 + 4r + 4 - r^2 = 8

4r + 4 = 8

4r = 4

r = 1

Thus, the radius of the larger circle is 1 + 2 = 3.

Answer: C
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Re: Two circles with radii r and r + 2 have areas that differ by 8π. What [#permalink]

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New post 09 Dec 2017, 00:35
Just plug in numbers. Obviously don't start with A or B. Start with D. Too big a difference, so answer is C.

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Re: Two circles with radii r and r + 2 have areas that differ by 8π. What   [#permalink] 09 Dec 2017, 00:35
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