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# In the figure above, each of the four large circles is

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In the figure above, each of the four large circles is [#permalink]  01 Oct 2020, 00:51
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Question Stats:

43% (01:50) correct 56% (01:53) wrong based on 16 sessions
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GRE In the figure above, each of the four large circles.png [ 11.14 KiB | Viewed 313 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
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Re: In the figure above, each of the four large circles is [#permalink]  01 Oct 2020, 01:10
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!

For more theory and practice see our GRE - Math Book
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Re: In the figure above, each of the four large circles is [#permalink]  01 Oct 2020, 03:19
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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
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GRE In the figure above, each of the four large circles.png [ 22.44 KiB | Viewed 289 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: 13673
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Kudos [?]: 3531 [0], given: 12627

Re: In the figure above, each of the four large circles is [#permalink]  01 Oct 2020, 09:09
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]  01 Oct 2020, 09:56
1
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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 259 times ]

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Re: In the figure above, each of the four large circles is   [#permalink] 01 Oct 2020, 09:56
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