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In the figure above, each of the four large circles is [#permalink]
Expert's post 00:00

Question Stats: 47% (01:52) correct 52% (01:53) wrong based on 17 sessions
Attachment: GRE In the figure above, each of the four large circles.png [ 11.14 KiB | Viewed 325 times ]

In the figure above, each of the four large circles is tangent to two of the other large circles, the small circle, and two sides of the square. If the radius of each of the large circles is 4, what is the diameter of the small circle?

A. $$\sqrt{2}$$

B. $$1$$

C. $$8 \sqrt{2} - 8$$

D. $$4 \sqrt{2} - 4$$

E. $$\sqrt{2} - 1$$
[Reveal] Spoiler: OA

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Founder  Joined: 18 Apr 2015
Posts: 13693
GRE 1: Q160 V160 Followers: 305

Kudos [?]: 3542 , given: 12654

Re: In the figure above, each of the four large circles is [#permalink]
Expert's post
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
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For more theory and practice see our GRE - Math Book
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- GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests. Moderator  Joined: 09 Jun 2020
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Re: In the figure above, each of the four large circles is [#permalink]
1
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as radius of large circle is 4 each side of square will be 8

diagonal of square is $$16 \sqrt{ 2 }$$

from the figure $$OA=4$$ and $$AB=4$$

OB=$$\sqrt{4^2 + 4^2}$$=$$4\sqrt{ 2 }$$

so radius of small circle will be =$$4\sqrt{ 2 }-4$$

diameter of small circle is $$8\sqrt{ 2 }-8$$

option C is correct
Attachments GRE In the figure above, each of the four large circles.png [ 22.44 KiB | Viewed 301 times ]

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Last edited by gajala on 01 Oct 2020, 19:56, edited 1 time in total.
Founder  Joined: 18 Apr 2015
Posts: 13693
GRE 1: Q160 V160 Followers: 305

Kudos [?]: 3542 , given: 12654

Re: In the figure above, each of the four large circles is [#permalink]
Expert's post
Great Explanation

Now look better, you forgot the tag
Code:
[m][/m] _________________

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Re: In the figure above, each of the four large circles is [#permalink]
1
KUDOS
Step 1: Understanding the question
Let A, B and O be the centers of the circles and radius of the smaller circle be x
As the diagonals of the square intersect at right angle, triangle AOB is a right angled at O

AB = 4+4 = 8
AO = (x+4)
BO = (x+4)

Applying Pythagoras
$$AB^2 = AO^2 + BO^2$$
64 = 2* $$(x+4)^2$$
32 = $$(x+4)^2$$
x + 4 = 4$$\sqrt{2 }$$
x = 4$$\sqrt{2}$$ - 4

Hence diameter of the smaller circle = 8$$\sqrt{2}$$ - 8
C is correct
Attachments GRE In the figure above, each of the four large circles.png [ 22.68 KiB | Viewed 271 times ]

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Want to Prepare for GRE Quant? Copy the below link: Re: In the figure above, each of the four large circles is   [#permalink] 01 Oct 2020, 09:56
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