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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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

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.

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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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