A square is inscribed in a circle. If ${ p }{ 1 }$ is the probability that a randomly chosen point of the circle lies within the square and ${ p }{ 2 }$ is the probability taht the point lies outside the square, then
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${ p }_{ 1 }={ p }_{ 2 }$
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${ p }_{ 1 }>{ p }_{ 2 }$ and $\displaystyle { { p }_{ 1 } }^{ 2 }-{ { p }_{ 2 } }^{ 2 }<\frac { 1 }{ 3 } $
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${ p }_{ 1 }<{ p }_{ 2 }$
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None of these
B
Correct answer
Explanation
Area of circle = pi*r^2. Side of inscribed square = r*sqrt(2), Area = 2*r^2. p1 = 2/pi approx 0.636. p2 = 1 - 2/pi approx 0.364. Clearly p1 > p2. p1^2 - p2^2 = (p1-p2)(p1+p2) = (4/pi - 1)(1) = 4/pi - 1 approx 1.27 - 1 = 0.27. Since 0.27 < 0.33, the condition holds.