The area of $\odot(O, r)$ is $240 \, cm^2$. In $\odot(O, r)$,minor arc $\widehat{ACB}$ subtends an angle of measure $45^\circ$ at the centre. Then,the area of minor sector $OACB$ is $\dots \dots \dots \, cm^2$.

In the figure,arcs are drawn by taking vertices $A, B$ and $C$ of an equilateral triangle of side $10 \, cm$ as centers to intersect the sides $BC, CA$ and $AB$ at their respective mid-points $D, E$ and $F$. Find the area of the shaded region (Use $\pi = 3.14$) (in $cm^2$).

In $\odot(O, r)$,$\overline{OA}$ and $\overline{OB}$ are two radii perpendicular to each other. If the perimeter of the minor sector formed by these radii is $20\,cm$,then $r = \ldots\,cm$.

The area of a circle is $75.46\, cm^{2}$. Find its circumference. (in $cm$)

As shown in the diagram, rectangle $ABCD$ is a metal sheet in which $CD = 20 \, cm$ and $BC = 14 \, cm$. From it, a semicircle with diameter $\overline{BC}$ and a sector with centre $A$ and radius $AD$ is cut off. Find the area of the remaining sheet in $cm^2$.

The radii of two concentric circles are $14 \, cm$ and $10.5 \, cm$. Then,the difference between their circumferences is $\ldots \ldots \ldots \, cm$.