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Viewed from the outside inward, the figure below depicts a square-circ

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Bunuel
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Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   05 Dec 2014, 07:38
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Tough and Tricky questions: Geometry.



Viewed from the outside inward, the figure below depicts a square-circle-square-circle, each enclosed within the other. If the area of square ABCD is 2 square units, then which of the following expresses the area of the darkened corners of square EFGH?
Attachment:
2014-12-05_1837.png
2014-12-05_1837.png [ 7.39 KiB | Viewed 5259 times ]


A) 2 − 1/4*π
B) 2 − 1/2*π
C) 1 − 1/4*π
D) 1/2− 1/8*π
E) 1 − 1/2*π

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C
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   05 Dec 2014, 10:53
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Whew... this problem just takes a while. Okay, starting with square ABCD, we know this has sides of length 2. That means that the first inscribed circle has a diameter of 2 (equal to the length of a side of ABCD). From this we can determine that the diameter of the first inscribed circle equals the diagonal of square EFGH (2), which means its sides are 2/sqrt(2). This yields that the second inscribed circle has a diameter of 2/sqrt(2) and a radius of 1/sqrt(2).

Now, the area of the shaded portions equals the difference between the area of square EFGH and the area of the circle inscribed within.
Area of square EFGH = side^2 = (2/sqrt(2))^2 = 4/2 = 2
Area of circle inscribed within EFGH = pi*r^2 = pi*(1/sqrt(2))^2 = pi/2
The difference between these areas = 2-pi/2

Choice B
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   05 Dec 2014, 10:59
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Area of square ABCD = 2
Therefore length of side AB = 2^0.5
Therefore length of diameter of big circle = 2^0.5
Therefore length of diagonal EG = 2^0.5
EFGH is a square. Therefore length of EF = 2^0.5/2^0.5 = 1
Therefore area of EFGH = 1
Length of diameter of small circle = 1
Therefore area of small circle = pi*(1/2)^2


Area of shaded region = 1-pi/4

Ans = C
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   06 Dec 2014, 00:28
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The bigger circle (inscribed within square ABCD) has a diameter equal to the diagonal of the square EFGH, whereas the small circle (inscribed within square EFGH) has a diameter equal to the side of the square EFGH. Thus, areas of the circles relate to each other: the area of the big inscribed circle is twice that of the small inscribed one and area of the big square ABCD is twice that of the small one EFGH.

It is given that side of square ABCD = 2 units

(Area of square ABCD) : (Area of Inscribed circle in square ABCD) : (Area of square EFGH): (Area of inscribed circle in square EFGH)

= \(4: \Pi : 2: \frac{\Pi}{2}\)

Area of darkened corners of EFGH
= difference of area between area of square EFGH and area of inscribed circle in square EFGH

= \(2 -\frac{\Pi}{2}\)

Answer : B
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   06 Dec 2014, 11:55
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intheend14 wrote:
Whew... this problem just takes a while. Okay, starting with square ABCD, we know this has sides of length 2. That means that the first inscribed circle has a diameter of 2 (equal to the length of a side of ABCD). From this we can determine that the diameter of the first inscribed circle equals the diagonal of square EFGH (2), which means its sides are 2/sqrt(2). This yields that the second inscribed circle has a diameter of 2/sqrt(2) and a radius of 1/sqrt(2).

Now, the area of the shaded portions equals the difference between the area of square EFGH and the area of the circle inscribed within.
Area of square EFGH = side^2 = (2/sqrt(2))^2 = 4/2 = 2
Area of circle inscribed within EFGH = pi*r^2 = pi*(1/sqrt(2))^2 = pi/2
The difference between these areas = 2-pi/2

Choice B



length of side of square ABCD will be root 2.

Area is given as 2 .


I think C should be the answer
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   06 Dec 2014, 11:57
Area of square ABCD = 2; side = root 2
diameter of big circle = root 2 so does the diagonal EG

EFGH is a square. length of EF = 1
diameter of small circle = 1
Therefore area of small circle = pi*(1/2)^2

Area of shaded region = 1-pi/4

C
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   06 Dec 2014, 18:40
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One side of the square is = Root 2. (2=S^2 ----> root 2=S)
Therefore the diameter of the big cicle is root 2, and so on the diagonal of the small square is root 2.
So one side of the small square is 1 (s root 2= root 2 ----> s= root 2/root 2 ----> s=1) and 1x1 is the area of the small square.
To find the area of the shaded region you have to substract the small square area (1) minus the small circle area that is (1/2) ^2
x pi. So the answer is 1 - 1/4 * pi. Answer C.
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   06 Dec 2014, 18:55
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Let outer Square and circle be S1 and C1 and inner squate and circle be S2 and C2.
Question: S2 - C2 ?

Given: S1 = 2

If a circle is inscribed in a square then the relationship between their areas is C = pi/4*S so C1 = pi/2
if a square is inscribed in a circle then the relationship between their areas is C = pi/2*S, therefore pi/2 = pi/2S2, S2 = 1

C2 = pi/4 therefore S2-C2 = 1-pi/4

Answer: C


Bunuel wrote:

Tough and Tricky questions: Geometry.



Viewed from the outside inward, the figure below depicts a square-circle-square-circle, each enclosed within the other. If the area of square ABCD is 2 square units, then which of the following expresses the area of the darkened corners of square EFGH?
Attachment:
2014-12-05_1837.png


A) 2 − 1/4*π
B) 2 − 1/2*π
C) 1 − 1/4*π
D) 1/2− 1/8*π
E) 1 − 1/2*π

Kudos for a correct solution.

Source: Chili Hot GMAT
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   08 Dec 2014, 06:08
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Bunuel wrote:

Tough and Tricky questions: Geometry.



Viewed from the outside inward, the figure below depicts a square-circle-square-circle, each enclosed within the other. If the area of square ABCD is 2 square units, then which of the following expresses the area of the darkened corners of square EFGH?
Attachment:
2014-12-05_1837.png


A) 2 − 1/4*π
B) 2 − 1/2*π
C) 1 − 1/4*π
D) 1/2− 1/8*π
E) 1 − 1/2*π

Kudos for a correct solution.

Source: Chili Hot GMAT


The correct answer is C.
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   08 Dec 2014, 21:13
Answer = C) 1 − 1/4*π

Refer diagram below:

Attachment:
2014-12-05_1837.png
2014-12-05_1837.png [ 13.8 KiB | Viewed 4753 times ]


Just rotate the square EFGH by 45 degrees & fade out the outer circle, you see that square EFGH is inscribed in square ABCD

So, area of square EFGH \(= \frac{1}{2} area of square ABCD = \frac{1}{2} * 2 = 1\)

Side of square EFGH = 1

Radius of the inscribed circle \(= \frac{1}{2}\)

Area of circle \(= \pi \frac{1}{2^2} = \frac{\pi}{4}\)

Area of shaded region \(= 1 - \frac{\pi}{4}\)
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink] New post   24 Oct 2023, 12:38
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Re: Viewed from the outside inward, the figure below depicts a square-circ [#permalink]
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