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

Question

https://gmatclub.com/forum/download/file.php?id=24540 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} 2014-10-28_2033.png

PareshGmat · Accepted Answer

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 rhom.png 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

SaharKhan · Answer

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

kishgau · Answer

Circumference = 2 * length of arc ADC = 2 * pi * sqrt(x)
radius of circle = sqrt (x)

Probability =   / area(circle) 
 = (pi - 2 )/ pi

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=vDtyMOcDlCU

JJo · Answer

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

adiagr · Answer

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

patriciadfer · Answer

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.

aritranandy8 · Answer

VeritasKarishma  plz explain

KarishmaB · Answer

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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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

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