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Each of the small circles in the figure has radius one. The innermost

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Each of the small circles in the figure has radius one. The innermost [#permalink] New post   Updated on: 18 Mar 2019, 03:39
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69% (01:26) correct 31% (01:39) wrong based on 29 sessions
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GMATbuster's Weekly Quant Quiz#8 Ques #5


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Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.

(A) \(\pi\)

(B) \(1.5\pi\)

(C) \(2\pi\)

(D) \(3\pi\)

(E) \(3.5\pi\)

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C

Originally posted by GMATBusters on 10 Nov 2018, 10:11.
Last edited by Bunuel on 18 Mar 2019, 03:39, edited 1 time in total.
Edited the question.
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:15
The diameter of the larger circle = diameter of 3 smaller circle = 6

Area of the shaded circle = Area of the bigger circle - Area of the 7 smaller circles
9Pi-7Pi
2Pi

C is the answer.
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:25
Area of large circle = Pi * 3 * 3
Area of all small circles = 7 * Pi * 1 *1

Ar of Large Circle - Total area of all small circles = 2Pi
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:26
Area of small circle is pi. Total area of 7 Small Circles is 7pi.
The diametre of big circle is 6. Radius 3. Area 9pi.

The difference is 9pi-7pi = 2pi
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:27
Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.


Area of shaded region= Area of larger- area of 6 smaller

Larger radius= if we join centres of 3 small circles, we get the diameter of the big circle.
Diameter= 2+2+2=6
Radius=3

So Shaded region= 9Pie-6*Pie

=3Pie


Hence Option D answer
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:52
Correct answer is C

The outer circle radius is 3pi. Area in 9pi

Inner circle area is pi
for 7 inner circles total area is 7pi

Shaded regio area = 9pi- 7pi= 2 pi .
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:58
pi*9 - pi*7 = 2*pi

Answer is Option C
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 10:59
Answer is C. 2Pi

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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 11:05
Diameter of bigger circle = 2+2+2 = 6
Radius of bigger circle = 3
Area of bigger circle = 9 pi
Smaller circle each area =pi
shaded area = bigger area - (7*(smaller area))
= 9 pi -7 pi =2pi
OPTION C
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink] New post   10 Nov 2018, 11:11
number of smaller circles with r=1 ::: 7
area of smaller circles = 7pi

radius of larger circle = radius of smaller circle+diameter of smaller circle= 1+2=3
area of larger circle :: 9pi

Area of shaded region= area of larger circle- area of smaller circles = 9pi-7pi=2pi
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Re: Each of the small circles in the figure has radius one. The innermost [#permalink]
10 Nov 2018, 11:11
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