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# If the inside circles are all of the same size and each

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If the inside circles are all of the same size and each [#permalink]

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04 Jun 2009, 12:24
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If the inside circles are all of the same size and each center is a corner of the interiour square, and the interiour circles lie on the circumference of the outside circle, what is the area of the of the highlighted green area?

(1) The circumference of the outside circle is $$5\Pi$$.
(2) The ratio of the outside circle's diameter to the radius of one of the interiour circles is $$2+\sqrt{2}:1$$.
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04 Jun 2009, 20:13
If the inside circles are all of the same size and each center is a corner of the interiour square, and the interiour circles lie on the circumference of the outside circle, what is the area of the of the highlighted green area?

(1) The circumference of the outside circle is $$5\Pi$$.
(2) The ratio of the outside circle's diameter to the radius of one of the interiour circles is $$2+\sqrt{2}:1$$.

A.
(1) The circumference of the outside circle is $$5\Pi$$.
large circle:
C = 2 PI R = 5 PI
R = 5/2
D = ac = ab+bf+cf = 5

Small & large circles:

redius of smal circles = ab = bd = de = cf.
If ab = x, bd = de = cf = x. Also be = ef = 2x
bf = sqrt(be^2 + ef^2)
bf = sqrt(4x^2 + 4x^2)
bf = sqrt(8x^2)
bf = 2x sqrt(2)

D = ab+bf+cf
5 = x + 2x sqrt(2) + x
5 = 2x + 2x sqrt(2)
5 = 2x [1+sqrt(2)]
x = 5/[2{1+sqrt(2)}]

Since we know diameters and redii of small and large circles and sides of the square, it is sufficient to find the area of the green colored space.

(2) The ratio doesnot help. Need absolute value. So insuff...
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Manager
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04 Jun 2009, 21:10
GMAT TIGER wrote:
If the inside circles are all of the same size and each center is a corner of the interiour square, and the interiour circles lie on the circumference of the outside circle, what is the area of the of the highlighted green area?

(1) The circumference of the outside circle is $$5\Pi$$.
(2) The ratio of the outside circle's diameter to the radius of one of the interiour circles is $$2+\sqrt{2}:1$$.

A.
(1) The circumference of the outside circle is $$5\Pi$$.
large circle:
C = 2 PI R = 5 PI
R = 5/2
D = ac = ab+bf+cf = 5

Small & large circles:

redius of smal circles = ab = bd = de = cf.
If ab = x, bd = de = cf = x. Also be = ef = 2x
bf = sqrt(be^2 + ef^2)
bf = sqrt(4x^2 + 4x^2)
bf = sqrt(8x^2)
bf = 2x sqrt(2)

D = ab+bf+cf
5 = x + 2x sqrt(2) + x
5 = 2x + 2x sqrt(2)
5 = 2x [1+sqrt(2)]
x = 5/[2{1+sqrt(2)}]

Since we know diameters and redii of small and large circles and sides of the square, it is sufficient to find the area of the green colored space.

(2) The ratio doesnot help. Need absolute value. So insuff...

Excellent work GMAT Tiger!
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04 Jun 2009, 21:36
I think we can do faster without calculation. Let's look at the figure. We have only one option how to build this system inside a circle (square cannot be lesser or larger, so too small circles). It means that all ratio between linear sizes are constant. So, if we know any linear size in the system, we can find other size and area.

1) suff. we know one linear size.
2) insuff. we don't know any linear size in the system.
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04 Jun 2009, 21:47
Yes that is the best way to approach it as well.

I would look at it and deduce that everything is locked into ratio-- there is only 1 way to to construct the shape. Hence we just need to know the length of something so we can scale it, and then thus answer question about it. Even rotating the circles inside the big circle doesn't change anything. (1) provides us with this

2 - Again it's a ratio, and doesn't give us a fixed value. Furthermore, since everything is locked into scale already, we can figure out any ratio from the shape. Hence insufficient

A
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09 Jun 2009, 01:32
Agree with A
Draw two perpendicular line which divide the 4 small circles into 4 equal parts and a line links the center of the the large and any small circle, we can see R = r + r sqrt 2.
From that we can calculate the green area = area of large circle - area of total 4 circles - (area of the square - area of a small circles)
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11 Jun 2009, 00:16
The correct answer is A. If (2) would involve an absolute value rather than a ratio it would be solvable and the answer would be D.

Pretty much everyone knows that you have to use an absolute value as input to get an absolute value as output (which is what they are asking for), thus even the 300-400 scorers will have a 50-50 chance on this one.
Re: Tough DS 12   [#permalink] 11 Jun 2009, 00:16
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