Let ABCD be a trapezium with parallel sides A | vedclass

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(D) Let $O$ be the center of the circle $S$ with diameter $AB$. Let $P$ be the point of tangency on $CD$. Since $AB \parallel CD$,the radius $OP$ is p

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$\frac{\pi}{6}$

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(D) Let $O$ be the center of the circle $S$ with diameter $AB$. Let $P$ be the point of tangency on $CD$. Since $AB \parallel CD$,the radius $OP$ is perpendicular to $CD$. Let $R$ be the mid-point of $AC$. Since $R$ lies on the circle with diameter $AB$,$\angle ARB = 90^{\circ}$. In $	riangle ABC$,$BR$ is the median to $AC$ and $BR \perp AC$,so $	riangle ABC$ is isosceles with $AB = BC$. Similarly,if $Q$ is the mid-point of $BD$,$\angle AQB = 90^{\circ}$,implying $	riangle ABD$ is isosceles with $AB = AD$. Thus,$AB = BC = AD$. Let $AB = 2r$,so the radius of the circle is $r$. The height of the trapezium is $r$. In the right-angled triangle formed by dropping a perpendicular from $A$ to $CD$ at $M$,we have $AM = r$ and $AD = AB = 2r$. Thus,$\sin(\angle ADM) = \frac{AM}{AD} = \frac{r}{2r} = \frac{1}{2}$. Therefore,$\angle ADM = 30^{\circ} = \frac{\pi}{6}$. The angles of the trapezium are $\frac{\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}, \frac{5\pi}{6}$. The smallest angle is $\frac{\pi}{6}$.

Let $ABCD$ be a trapezium with parallel sides $AB$ and $CD$ such that the circle $S$ with $AB$ as its diameter touches $CD$. Further,the circle $S$ passes through the mid-points of the diagonals $AC$ and $BD$ of the trapezium. The smallest angle of the trapezium is

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