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#### karush

##### Well-known member

- Jan 31, 2012

- 2,997

- Thread starter karush
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You have three circles inside. Call the radius of the largest inner circle r. Then the radius of the two smaller circles are r/2 and the radius of the big circle on the outside is 2r.Ok this is considered a "hard" GRE geometry question... notice there are no dimensions

How would you solve this in the fewest steps?

You are supposed to give the answer as a ratio so the r's eventually cancel out.

Can you finish?

-Dan

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- Jan 31, 2012

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Kinds I know that the area of a circle isYou have three circles inside. Call the radius of the largest inner circle r. Then the radius of the two smaller circles are r/2 and the radius of the big circle on the outside is 2r.

You are supposed to give the answer as a ratio so the r's eventually cancel out.

Can you finish?

-Dan

$A=\pi r^2$

- Mar 1, 2012

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Come on ... you can do this

Big circle - (medium circle + 2 small circles)

Big circle - (medium circle + 2 small circles)

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$\dfrac{\pi(1)^2}{\pi(2)^2}=\dfrac{1}{4}$

Kinda...like..

- Aug 30, 2012

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Check those radii again... Try calling the radius of one of the small circles to be r = 1 and give it another try. Then look at this:

$\dfrac{\pi(1)^2}{\pi(2)^2}=\dfrac{1}{4}$

Kinda...like..

Area of the two smaller circles: \(\displaystyle 2a = 2 ( \pi (r/2)^2 ) = (1/2) \pi r^2\)

Area of the whole circle: \(\displaystyle A_{big} = \pi (2r)^2 = 4 \pi r^2\)

Sum of the area of the inner circles: \(\displaystyle \pi r^2 + (1/2) \pi r^2 = (3/2) \pi r^2\)

So what is the ratio?

-Dan

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- Jan 31, 2012

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Check those radii again... Try calling the radius of one of the small circles to be r = 1 and give it another try. Then look at this:

Area of large circle in the center: \(\displaystyle A =\pi r^2\)

Area of the two smaller circles: \(\displaystyle 2a = 2 ( \pi (r/2)^2 ) = (1/2) \pi r^2\)

Area of the whole circle: \(\displaystyle A_{big} = \pi (2r)^2 = 4 \pi r^2\)

Sum of the area of the inner circles: \(\displaystyle \pi r^2 + (1/2) \pi r^2 = (3/2) \pi r^2\)

$\dfrac{4 \pi r^2-(3/2) \pi r^2}{4 \pi r^2}=\dfrac{8 \pi r^2-(3) \pi r^2}{8 \pi r^2}=\dfrac{5}{8}$

maybe

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- Mar 1, 2012

- 977

not maybe ... 5/8 is correct

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- Jan 31, 2012

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mahalo sorry my likes were so late

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