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If the figure below is a square with a side of 4 units, what is the ar

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If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 24 Nov 2014, 07:23
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If the figure below is a square with a side of 4 units, what is the area of the enclosed circle, expressed to the nearest whole number?
Image

A) π
B) 4
C) 8
D) 13
E) 16

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Source: Chili Hot GMAT

Attachment:
2014-11-24_1823.png
2014-11-24_1823.png [ 3.4 KiB | Viewed 1878 times ]

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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 24 Nov 2014, 07:40
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If the length of the side of the square is 4 units, the radius of the circle must be 2 units. Therefore, the area of the circle is πr^2 = π(4) = 3.1415 * 4 = approx. 13 units.

Answer is D.
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 24 Nov 2014, 12:47
1
Side of the square is \(4\) units.
Diameter of circle = side of square =\(4\)
Radius(r) of circle = \(\frac{Diameter}{2}\) =\(\frac{4}{2}\) =2
Area of circle = pi\(r^2\) --PS: i don't know how to insert π notation so i used pi.--

Area of circle= \(3.14*2^2\) \(=12.56\)
Area is approx \(13\) unit

Answer: D
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 24 Nov 2014, 20:19
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Answer = D) 13

Circle is inscribed, so radius = 2

\(Area = \pi 2^2 = 3.14 * 4 = 12.56 \approx{13}\)
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 25 Nov 2014, 04:51
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From the figure the diameter of the circle =4 units

So area = Area = 3.15*r^2=3.15*4 =13 Approx.


Answer D
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 25 Nov 2014, 05:27
1
Diameter of the circle = Side of the square = 4 units

Area of the circle = 3.14 * (Diameter)^2/4 = 3.14* 16/4 = 12.56 = 13 (Approx)

Answer D.
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 25 Nov 2014, 07:53
Bunuel wrote:
If the figure below is a square with a side of 4 units, what is the area of the enclosed circle, expressed to the nearest whole number?
Image
A) π
B) 4
C) 8
D) 13
E) 16

Kudos for a correct solution.

Source: Chili Hot GMAT


Official Solution:

If the figure below is a square with a side of 4 units, what is the area of the enclosed circle, expressed to the nearest whole number?
Image

A) π
B) 4
C) 8
D) 13
E) 16

Eyeballing:
Eyeballing is a parallel technique to be used on diagrams. Note that whatever the area of this circle may be, it must be less than the area of this square. The area of the square (in square units) is: A = s^2 = 4 × 4 = 16. Therefore, the area of the circle is a little less than 16. Choice D is the only close answer. For the record, the near exact area of the circle is: A = π^2 = 3.14(2)^2 = 12.56 or 13. Note that the decimal approximation for π is 3.14 while the fractional approximation is 22/7.

Answer: D.
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 23 Jan 2018, 09:01
Bunuel wrote:
If the figure below is a square with a side of 4 units, what is the area of the enclosed circle, expressed to the nearest whole number?
Attachment:
2014-11-24_1823.png


A) π
B) 4
C) 8
D) 13
E) 16


Since the square has a side of 4, the diameter of the circle is 4, and the radius is 2.

Thus, the area is π x 2^2 = 4π ≈ 4 x 3.14 = 12.56. The nearest integer to 12.56 is 13.

Answer: D
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Re: If the figure below is a square with a side of 4 units, what is the ar  [#permalink]

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New post 23 Jan 2018, 10:19
Bunuel wrote:
If the figure below is a square with a side of 4 units, what is the area of the enclosed circle, expressed to the nearest whole number?
Image

A) π
B) 4
C) 8
D) 13
E) 16

Kudos for a correct solution.

Source: Chili Hot GMAT

Attachment:
2014-11-24_1823.png


Radius of the circle is 2 units.

Area of the circle is \(π2^2\) = \(4π\) ~ \(13\) , Answer will be (D)

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Re: If the figure below is a square with a side of 4 units, what is the ar &nbs [#permalink] 23 Jan 2018, 10:19
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