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605-655 Level|   Geometry|   Probability|      
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Bunuel
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ABCD is a square inscribed in a circle and arc ADC has a length of 28 Oct 2014, 09:35
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45% (medium)

Question Stats:

73% (02:57) correct 27% (02:53) wrong based on 595 sessions

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ABCD is a square inscribed in a circle and arc ADC has a length of \(\pi\sqrt{x}\). If a dart is thrown and lands somewhere in the circle, what is the probability that it will not fall within the inscribed square? (Assume that the point in the circle where the dart lands is completely random.)

(A) \(2x\)

(B) \(π(x) - 2x\)

(C) \(π(x) - \sqrt{2}(x)\)

(D) \(1 - \frac{2}{π}\)

(E) \(1 - \frac{2}{x}\)

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Attachment:
2014-10-28_2033.png
2014-10-28_2033.png [ 10.59 KiB | Viewed 26823 times ]
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PareshGmat
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ABCD is a square inscribed in a circle and arc ADC has a length of Updated on: 08 Dec 2014, 03:47
Originally posted by PareshGmat on 28 Oct 2014, 21:41.
Last edited by PareshGmat on 08 Dec 2014, 03:47, edited 2 times in total.
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Perimeter of the circle \(= 2 * \pi\sqrt{x}\)

So, radius of circle \(= \sqrt{x}\)

Diameter of the circle = Diagonal of the square \(= 2\sqrt{x}\)

Area of circle \(= \pi (\sqrt{x})^2 = \pi x\)

Area of square \(= \frac{(2\sqrt{x})^2}{2} = 2x\)

Area of the yellow shaded region \(= \pi x - 2x\)
Attachment:
rhom.png
rhom.png [ 11.37 KiB | Viewed 25112 times ]

Probability of NOT falling in inscribed square = Probability of FALLING in the shaded region

Probability \(= \frac{\pi x - 2x}{\pi x} = 1 - \frac{2}{\pi}\)

Bunuel: Kindly update the OA
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ABCD is a square inscribed in a circle and arc ADC has a length of 28 Oct 2014, 10:36
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Length of the arc = pi*sqrt x
Radius of the circle = 180/360*2pi*r = pi*sqrt x ==> r = sqrt x
Area of the circle = pi*r^2 ==> pi*x

Diagonal of the square = diameter of the circle = 2* sqrt x
Side(S) of the square =Diagonal/sqrt 2
==> S * sqrt 2 = 2*sqrt x ==> s=sqrt 2x
Area of the square = s^2 = 2x

Probability of the dart not landing on the square = 1-2x/pi*x ==> 1- 2/pi

:-D
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ABCD is a square inscribed in a circle and arc ADC has a length of 28 Oct 2014, 11:56
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Circumference = 2 * length of arc ADC = 2 * pi * sqrt(x)
radius of circle = sqrt (x)

Probability = [area(circle) - area(square)] / area(circle)
= (pi - 2 )/ pi
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ABCD is a square inscribed in a circle and arc ADC has a length of Updated on: 27 Nov 2021, 09:01
Originally posted by BrushMyQuant on 06 Dec 2014, 23:39.
Last edited by BrushMyQuant on 27 Nov 2021, 09:01, edited 1 time in total.
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Hi Paresh,

Area of yellow shaded region is \(= \pi x - 2x\) (circle - square) and not \(= 2x - \pi x\) (square - circle)

kindly update the remaining steps too
Thank you.

Watch the following video to Learn Basics of Circles




PareshGmat
Area of the yellow shaded region \(= 2x - \pi x\)
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ABCD is a square inscribed in a circle and arc ADC has a length of 07 Dec 2014, 22:23
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PareshGmat
nktdotgupta
Hi Paresh,

Area of yellow shaded region is \(= \pi x - 2x\) (circle - square) and not \(= 2x - \pi x\) (square - circle)

kindly update the remaining steps too
Thank you.
PareshGmat
Area of the yellow shaded region \(= 2x - \pi x\)

Its been updated. Thanks

Is there any way to update OA for this question?

Hi Paresh,

I believe the denominator of the probability should be area of Circle and not that of square.

So, the prob (of not falling in inscribed square) = 1 - P(falling in inscribed square) = 1 - (2*x/pi*x) = 1-2/pi

OR as per your approach prob (of not falling in inscribed square) = (pi*x - 2x) / pi*x = 1-2/pi
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ABCD is a square inscribed in a circle and arc ADC has a length of 27 May 2016, 06:37
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First Point Arc ADC is nothing but semi circle. Because AC is diameter.

so length of arc ADC is (\(\pi\) r)

Given. (\(\pi\) r = (\(\pi\) \(\sqrt{x}\)

r = \(\sqrt{x}\)

Area of the Circle: \(\pi\) \(r^2\) = \(\pi\)x

Now let us calculate area of square.

Diagonal of square = 2r = 2\(\sqrt{x}\)

a \(\sqrt{2}\) = 2\(\sqrt{x}\); where a is the side of square.

a = \(\sqrt{2x}\)

Area of square = 2x.

We are asked probability that dart will not fall within the inscribed square?

Probability = (\(\pi\)x - 2x) / \(\pi\)x = (\(\pi\) - 2) / \(\pi\)
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ABCD is a square inscribed in a circle and arc ADC has a length of 25 Jan 2018, 11:45
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A very simple way to solve this question is by looking the alternatives. The probability of the dard fall outside the square does not depend on the size of the circle and the square, thus it does not depend on x. The only alternative that does not depend on x is letter d.
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ABCD is a square inscribed in a circle and arc ADC has a length of 23 Feb 2021, 04:37
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VeritasKarishma plz explain
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ABCD is a square inscribed in a circle and arc ADC has a length of 23 Feb 2021, 23:59
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Bunuel

ABCD is a square inscribed in a circle and arc ADC has a length of \(\pi\sqrt{x}\). If a dart is thrown and lands somewhere in the circle, what is the probability that it will not fall within the inscribed square? (Assume that the point in the circle where the dart lands is completely random.)

(A) \(2x\)

(B) \(π(x) - 2x\)

(C) \(π(x) - \sqrt{2}(x)\)

(D) \(1 - \frac{2}{π}\)

(E) \(1 - \frac{2}{x}\)

Show Spoiler
Attachment:
2014-10-28_2033.png

The thing to note here is that since the square is inscribed in the circle, its diagonal will be the circle's diameter. Thereafter, the question becomes quite simple since getting any one dimension of these symmetrical figures is enough to get all others.

See: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/0 ... n-circles/

Now if AC is the diagonal and the diameter, it means ADC is a semi circle.

Arc ADC = \(\pi*r\) which means \(r = \sqrt{x}\)

Area of circle = \(\pi*r^2 = \pi*\sqrt{x}^2 = \pi*x\)

Area of square = \(side^2 = (Diagonal/\sqrt{2})^2 = (2\sqrt{x}/\sqrt{2})^2 = 2x\)

Area of square/ Area of Circle \(= 2/\pi\)

Probability that it will land outside the square \(= 1 - 2/\pi\)

Answer (D)
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