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# The above picture depicts a circle perfectly inscribed in a square. Is

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Math Expert
Joined: 02 Sep 2009
Posts: 51280
The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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25 Aug 2015, 23:46
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:12) correct 30% (02:01) wrong based on 122 sessions

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The above picture depicts a circle perfectly inscribed in a square. Is the shaded region > 4?

(1) The area of the large rectangle equals 64.
(2) The perimeter of the shaded region equals 8 + 2π.

Kudos for a correct solution.

Attachment:

circle-in-square-red2.GIF [ 2.24 KiB | Viewed 1558 times ]

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Joined: 14 Mar 2014
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GMAT 1: 710 Q50 V34
The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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26 Aug 2015, 08:59
IMO: D

Consider the following fig:
Attachment:

Capture.JPG [ 15.19 KiB | Viewed 1388 times ]

Question Stem: Area of the shaded region = $$\frac{[Area of the square - Area of the cirlce]}{4}$$
Area of square = 2r*2r = $$4r^2$$
Area of Circle = $$πr^2$$
Area of the shaded region = $$\frac{[4r^2 - πr^2]}{4}$$

IS $$\frac{r^2[4 - π]}{4}$$ > 4 ?

IS $$r^2[4 - π]$$ > 16 ? -- (i)

St 1: The area of the large rectangle equals 64.

I think the question is flawed. PSTU is a rectangle and PQRS is also a rectangle(Every square is a rectangle). The use of word large is ambiguous. It can either represent PSTU or PQRS. If largest is used instead then PQRS can be considered which would be sufficient alone as well.

Area of PSTU = 64
Where PS = 2r & SU = r
Thus, 2r*r = 64
$$r^2$$= 32 ---(ii)

Sub in Eq - (i)
$$32[4 - π]$$ > 16 ? YES
Hence Suff

St 2: The perimeter of the shaded region equals 8 + 2π

Perimeter of the shaded region = Length of the arc TV + Side TQ + Side QV

Length of the arc = $$\frac{θ}{360} * 2πr$$
= $$\frac{90}{360} * 2πr$$
= $$\frac{πr}{2}$$

Thus,

$$8 + 2π = \frac{πr}{2} + r + r$$
r = 4
Sub in eq--(i)
$$r^2[4 - π]$$ > 16 ?
$$16[4 - π]$$ > 16 ? YES
Hence Suff
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Re: The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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26 Aug 2015, 18:11
2
If we can determine the radius, we can determine the side of the square and the rectangle, allowing us to find the area of the square, rectangle and circle. What is the radius?

(1) The area of the large rectangle equals 64.
r*2r=64
Sufficient

(2) The perimeter of the shaded region equals 8 + 2π.
8 + 2π = (2πr)/4 + 2r
Sufficient

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Posts: 1919
Re: The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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27 Aug 2015, 03:33
Hi Bunuel:
In statement 1: r=4\sqrt{2}
In Statement 2: r =4
Clearly, the question is flawed or do I miss something?
Math Expert
Joined: 02 Sep 2009
Posts: 51280
Re: The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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30 Aug 2015, 08:33
Bunuel wrote:

The above picture depicts a circle perfectly inscribed in a square. Is the shaded region > 4?

(1) The area of the large rectangle equals 64.
(2) The perimeter of the shaded region equals 8 + 2π.

Kudos for a correct solution.

Attachment:
circle-in-square-red2.GIF

GROCKIT OFFICIAL SOLUTION:

Statement 1: By knowing the area of the large square we also know the lengths of its sides. (Note that 64 is a perfect square, which should be a clue.) If the side is 8, then so is the diameter, which means the radius equals 4. In the image, we can see that the “larger figure” is the top-right square bordered by two radii and the outer border. How do we know that it’s a square? Two pieces of information: All sides are equal to 4, and the radius meets the large square at a right angle because it is a tangent. In this instance, the area of the smaller square equals 16. Since the interior angle is 90-degrees (360/4), the area of the sector of the circle can be represented by A = πr²/4. So that A = 16π/4 = 4π.

A(shaded) = A(small square) – A(sector) = 16 – 4π < 4 because 4π > 12. Sufficient.

Statement 2: The first thing that should jump out is the combination of a π-term and non- π-term. We can reasonably assume that 8 represents the two straight sides of the perimeter and the 2π the arc-length of the quarter circle, which is known because of the internal right angle. If 2π = C/4, then C = 8π. If C = 8π, then r = 4. From here, we return to the same reasoning as above:

A(shaded) = A(small square) – A(sector) = 16 – 4π < 4 because 4π > 12. Sufficient.

Each statement is sufficient, so the answer is choice D.
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Re: The above picture depicts a circle perfectly inscribed in a square. Is  [#permalink]

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12 Jul 2018, 11:01
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Re: The above picture depicts a circle perfectly inscribed in a square. Is &nbs [#permalink] 12 Jul 2018, 11:01
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