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In the figure shown above, $\text{PQRS}$ is a square. The shaded portion is formed by the intersection of sectors of circles with radius equal to the side of the square and centers at $S$ and $Q$.

The probability that any point picked randomly within the square falls in the shaded area is ____________

1. $4-\frac{\pi }{2}$
2. $\frac{1}{2}$
3. $\frac{\pi }{2}-1$
4. $\frac{\pi }{4}$
Migrated from GO Civil 7 months ago by Arjun

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Sector PRS area = Sector QPR Area = π$r^{2}$/4

Required area = (sum of sectors PRS and QPR area) – Square PQRS area

=  $r^{2}$/4  –$r^{2}$

Probability = Favourable area / Total area

= ($r^{2}$(π/2  –1)) / $r^{2}$

= π/2  –1

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Let the side of the square be $1\;\text{cm},$ then $r = 1\;\text{cm}.$

From the center $S,$ we can draw the circle, and only the quarter part is shown in the figure.

The shaded area $A_{1} = A_{\text{square}} – A_{\text{quarter circle}} = 1 – \dfrac{\pi}{4}$

From the center $Q,$ we can draw the circle, and only the quarter part is shown in the figure.

The shaded area $A_{2} = A_{\text{square}} – A_{\text{quarter circle}} = 1 – \dfrac{\pi}{4}$

Now, we can combine the above two figures, we get,

Now, the shaded area $= A_{\text{square}} – A_{1} – A_{2}$

$\quad = 1 – \left(1-\dfrac{\pi}{4}\right) – \left(1-\dfrac{\pi}{4}\right)$

$\quad = – 1 + \dfrac{2\pi}{4} = \dfrac{\pi}{2} – 1$

Now, the required probability $= \dfrac{\text{Favorable shaded area}}{\text{Total area}} = \dfrac{\frac{\pi}{2} – 1}{1} = \dfrac{\pi}{2} – 1.$

$\therefore$ The probability that any point picked randomly within the square falls in the shaded area $= \dfrac{\pi}{2} – 1.$

So, the correct answer is $(C).$

Similarly, do for another side

Therefore, Area of shaded area = $2 * (\frac{ r^{2}}{4}(\pi – 2)) = \frac{ r^{2}}{2}(\pi – 2)$

Now, total area = $r^{2}$

Therefore, Probability$\frac{ \frac{ r^{2}}{2}(\pi – 2)}{r^{2}}$

= $\frac{ 1}{2}(\pi – 2)$

= $\frac{ \pi}{2} – 1$

So, Ans is $C$

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