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The area of the shaded region is 12, and the sum of the length of a si

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Math Expert
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Joined: 02 Sep 2009
Posts: 56357
The area of the shaded region is 12, and the sum of the length of a si  [#permalink]

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New post 23 Aug 2018, 03:19
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

100% (01:53) correct 0% (00:00) wrong based on 15 sessions

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Status: Learning stage
Joined: 01 Oct 2017
Posts: 1028
WE: Supply Chain Management (Energy and Utilities)
Re: The area of the shaded region is 12, and the sum of the length of a si  [#permalink]

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New post 23 Aug 2018, 03:29
Bunuel wrote:
Image
The area of the shaded region is 12, and the sum of the length of a side of the outer square and a side of the inner square is 6. What is the perimeter of the outer square?


A. 48
B. 24
C. 16
D. 12
E. 8


Attachment:
image017.jpg


Let 'A' and 'a' are the length of sides outer square and inner square respectively.
Given, sum of the length of a side of the outer square and a side of the inner square is 6.
Or, A+a=6---(1)
Given, area of the shaded region is 12
Or, \(A^2-a^2=12\)
Or, (A+a)(A-a)=12
Or, 6*(A-a)=12
Or, A-a=2---(2)

Adding (1) and (2), we have 2A=8 Or, A=4

So, perimeter of outer square=4A=4*4=16

Ans. (C)
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Re: The area of the shaded region is 12, and the sum of the length of a si   [#permalink] 23 Aug 2018, 03:29
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The area of the shaded region is 12, and the sum of the length of a si

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