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In the figure above, SAND and SURF are squares, and O is the center of

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In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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New post Updated on: 26 May 2017, 05:02
5
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

58% (09:48) correct 42% (03:04) wrong based on 74 sessions

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In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1



Attachment:
Untitled.png
Untitled.png [ 36.59 KiB | Viewed 1361 times ]

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Originally posted by hazelnut on 26 May 2017, 04:02.
Last edited by Bunuel on 26 May 2017, 05:02, edited 2 times in total.
Edited the question.
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Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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New post 26 May 2017, 05:18
2
hazelnut wrote:
Image
In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1

Attachment:
Untitled.png


Side of SAND – x

Side of SURF – y

Radius of the circle – r

Now let’s take a look at the right triangle SUA

x^2 + y^2 = 4r^2

r^2 = (x^2 + y^2)/4

Taking the required fraction:

C/Q = π*r^2 / (x^2 + y^2) = π/4 = 3.14/4 is slightly more than 3/4.

Answer C.
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Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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New post 26 May 2017, 05:43
hazelnut wrote:
Image
In the figure above, SAND and SURF are squares, and O is the center of the circle. If Q is the sum of the areas of squares SAND and SURF and C is the area of the circle, then the fraction C/Q is

(A) less than 5/8

(B) between 5/8 and 3/4

(C) between 3/4 and 7/8

(D) between 7/8 and 1

(E) greater 1



Attachment:
Untitled.png



Let side of SAND be "a" and SURF be "b"

AREA of SAND & SURF will be \(a^2+b^2\)

Radius of circle = \sqrt{a^2 + b^2}
On solving
Area of circle/Area of Squares = 3.14/4 which lies between 3/4 & 7/8 hence option C
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Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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New post 26 May 2017, 23:33
Let's take radius of circle=x
Then Q= 4(x )square
C= 2*22/7*x*x
So C/Q= 11/7


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Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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New post 27 May 2017, 05:32
Since Triangle USA is a right angled triangle with UA as diameter of the circle.

If we assume sides of sqaure as 6 and 8 respectively
Then 6^2 + 8^2 = 10^2 = UA^2

Which is basically sum of the areas of two sqaures I.e. P

Area of circle C =( π•10^2 )/4

Ratio C /P = [ ( π•10^2 )/4 ] ÷ 10^2
= π / 4
= 3.14 / 4 , greater than 3/4

Option C

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Re: In the figure above, SAND and SURF are squares, and O is the center of  [#permalink]

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Re: In the figure above, SAND and SURF are squares, and O is the center of &nbs [#permalink] 04 Nov 2018, 21:11
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