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Bunuel
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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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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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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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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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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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VeritasKarishma plz explain
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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) \(1 - \frac{2}{x}\)

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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