In $\odot(O, 12)$, minor $\widehat{ACB}$ subtends an angle of measure $30^{\circ}$ at the centre. Then, the length of major $\widehat{ADB}$ is $\ldots \ldots \ldots \text{ cm}$. (in $\pi$)

In $\odot(O, 5.6)$,$\overline{OA}$ and $\overline{OB}$ are radii perpendicular to each other. Then,the difference between the area of the minor sector formed by minor $\widehat{AB}$ and the corresponding minor segment is $\ldots \ldots \ldots \ldots cm^2$.

Find the area of the shaded region in the figure where arcs drawn with centers $A, B, C,$ and $D$ intersect in pairs at mid-points $P, Q, R,$ and $S$ of the sides $AB, BC, CD,$ and $DA,$ respectively of a square $ABCD$ (Use $\pi = 3.14$) (in $cm^{2}$).

The length of the minute hand of a clock is $12\,cm$. The area of the region swept by it in $5$ minutes is $\ldots \ldots \ldots \,cm^2$. $(\pi=3.14)$

Is it true that the distance travelled by a circular wheel of diameter $d \text{ cm}$ in one revolution is $2 \pi d \text{ cm}$? Why?

The area of a circle is $38.5\, m^2$,then its circumference will be $\ldots \ldots \ldots \ldots \, m$.