In the figure,ABCD is a unit square. A circle | vedclass

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(D) Let the side of the square $ABCD$ be $1$. Thus,$AB = BC = CD = DA = 1$.Since $O$ lies on the extension of $CD$,let $OD = x$. Then $OC = x + 1$.In

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$\frac{6-\pi}{4}$

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(D) Let the side of the square $ABCD$ be $1$. Thus,$AB = BC = CD = DA = 1$.Since $O$ lies on the extension of $CD$,let $OD = x$. Then $OC = x + 1$.In $	riangle OAC$,$AC$ is tangent to the circle at $A$,so $\angle OAC = 90^{\circ}$.In $	riangle ADC$,$\angle DAC = 45^{\circ}$. Since $\angle OAC = 90^{\circ}$,$\angle OAD = 90^{\circ} - 45^{\circ} = 45^{\circ}$.In $	riangle OAD$,$	an(45^{\circ}) = \frac{OD}{AD} \implies 1 = \frac{x}{1} \implies x = 1$.Thus,$OD = 1$ and the radius $R = OA = \sqrt{AD^2 + OD^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.The shaded region consists of the square $ABCD$ minus the area of the circular sector $AXD$ (where $X$ is the intersection of the circle with $CD$).Area of square $ABCD = 1^2 = 1$.The sector $OAX$ has radius $R = \sqrt{2}$ and central angle $\angle AOX = 90^{\circ} - 45^{\circ} = 45^{\circ}$.Area of sector $OAX = \frac{45}{360} 	imes \pi 	imes R^2 = \frac{1}{8} 	imes \pi 	imes 2 = \frac{\pi}{4}$.Area of $	riangle OAD = \frac{1}{2} 	imes AD 	imes OD = \frac{1}{2} 	imes 1 	imes 1 = 0.5$.Area of shaded region = Area of square $ABCD$ - (Area of sector $OAX$ - Area of $	riangle OAD$) = $1 - (\frac{\pi}{4} - 0.5) = 1.5 - \frac{\pi}{4} = \frac{6-\pi}{4}$.

In the figure,$ABCD$ is a unit square. $A$ circle is drawn with centre $O$ on the extended line $CD$ and passing through $A$. If the diagonal $AC$ is tangent to the circle,then the area of the shaded region is

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