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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, 06:02
10
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

54% (02:49) correct 46% (03:06) wrong based on 101 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 1883 times ]

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Originally posted by hazelnut on 26 May 2017, 05:02.
Last edited by Bunuel on 26 May 2017, 06:02, 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, 06:18
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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, 06:43
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 27 May 2017, 00:33
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, 06:32
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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New post 24 Aug 2019, 08:56
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


I have a slightly different approach.
You can see the right triangle. It is right there. So why not to just assume the sides? Let's say it is 6,8,10 Triplet.
Area of squares is 64+36 which is 100.
Area of Circle is Pi*5*5 which works out to be approximately 75. Since you know actual value is higher the fraction will be higher than 3/4. Hence C.
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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 25 Aug 2019, 02:59
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


Given: In the figure above, SAND and SURF are squares, and O is the center of the circle.

Asked: 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

\(Q = a^b + b^2 = D^2\) where D is the diameter of the circle

\(C = \pi r^2 = \frac{\pi D^2}{4} = \frac{\pi}{4} Q\)

\(\frac{C}{Q} = \frac{\pi}{4} =\frac{7}{8} > \frac{3.14}{4} > \frac{3}{4}\)

IMO C
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In the figure above, SAND and SURF are squares, and O is the center of   [#permalink] 25 Aug 2019, 02:59
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