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# ABCD is a square inscribed in a circle and arc ADC has a length of

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ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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28 Oct 2014, 09:35
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Difficulty:

55% (hard)

Question Stats:

64% (03:05) correct 36% (02:29) wrong based on 213 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) $$1 - \frac{2}{x}$$

Attachment:

2014-10-28_2033.png [ 10.59 KiB | Viewed 8060 times ]

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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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Updated on: 08 Dec 2014, 03:47
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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 [ 11.37 KiB | Viewed 7561 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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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.
##### General Discussion
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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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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

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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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28 Oct 2014, 11:56
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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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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06 Dec 2014, 23:39
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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.
PareshGmat wrote:
Area of the yellow shaded region $$= 2x - \pi x$$

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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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07 Dec 2014, 22:23
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PareshGmat wrote:
nktdotgupta wrote:
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 wrote:
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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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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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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Re: ABCD is a square inscribed in a circle and arc ADC has a length of  [#permalink]

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25 Jan 2018, 11:45
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.
Re: ABCD is a square inscribed in a circle and arc ADC has a length of &nbs [#permalink] 25 Jan 2018, 11:45
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