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ABCD is a square inscribed in a circle and arc ADC has a length of
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17 Mar 2021, 03:07
17 Mar 2021, 03:07
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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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ABCD is a square inscribed in a circle and arc ADC has a length of
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17 Mar 2021, 08:08
17 Mar 2021, 08:08
1
GeminiHeat wrote:
Attachment:
2014-10-28_2033.png
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) \(1 - \frac{2}{x}\)
Since, Arc ADC = \(\pi\sqrt{x} = \frac{180}{360}(2)(\pi)(r)\)
\(r = \sqrt{x}\)
Now, the required probability = [Area of Circle - Area of Square] / Area of Circle[/fraction][/m]
Area of Circle = \(\pi(\sqrt{x})^2 = \pi(x)\)
To find the Area of Square, we must find the it's side first!
Notice, Diagonal of the square is the diameter of the circle
So, \(\sqrt{2}s = 2r\)
\(s = \sqrt{2}r\)
Area of square = \(s^2 = 2r^2 = 2(\sqrt{x})^2 = 2x\)
Probability = \(\frac{{\pi(x) - 2x}}{{\pi(x)}} = 1 - \frac{2}{\pi}\)
Hence, option D
Re: ABCD is a square inscribed in a circle and arc ADC has a length of
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26 Oct 2022, 08:28
26 Oct 2022, 08:28
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