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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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Updated on: 26 May 2017, 05:02
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75% (hard)

Question Stats:

56% (02:44) correct 44% (03:07) wrong based on 108 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 [ 36.59 KiB | Viewed 2106 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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26 May 2017, 05:18
2
hazelnut wrote:

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.

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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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26 May 2017, 05:43
1
hazelnut wrote:

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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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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27 May 2017, 05:32
1
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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24 Aug 2019, 07:56
hazelnut wrote:

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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25 Aug 2019, 01:59
hazelnut wrote:

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, 01:59

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

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