AB and CD are arcs of the two concentric circles of radii 21cm and 7cm and centre ‘O’ if \[\angle \text{AOB}={{30}^{\circ }}\]. Find the area of the shaded region.

Hint: The area of the shaded region can be found by subtracting the area of sector COD from the area of sector AOB. Subtract the value of \[\theta \] and radius and find the area of sector AOB and COD.

Complete step-by-step solution
We have been given the sector of two concentric circles.
The center of the circle is given 0. Now as it is said that these are the area of 2 concentric circles, then their radii of the circle can be given as, from figure
OB = 21cm, radii of outer circle.
OD = 7cm, radii of inner circle.
Thus we need to find the shaded region’s area. That is we need to find the area of ABCD

Thus we can say that,
Area of shaded region = Area of sector AOB – area of sector COD \[\to \left( 1 \right)\]
Now let us first find the area of sector AOB.
From figure, we can say that \[\theta ={{30}^{\circ }}\] and radius, OB = 21cm.
We know that the area of a sector is given by,
\[area\text{ }=\dfrac{\theta }{360}\times \text{ }\pi {{r}^{2}}\]
\[\therefore Area\text{ }of\sec tor\text{ }AOB\text{ }=\dfrac{\theta }{360}\times \text{ }\pi {{r}^{2}}\]
 \[\begin{align}
  & =\dfrac{30}{360}\times \text{ }\pi \times {{\left( 21 \right)}^{2}} \\
 & =\dfrac{1}{12}\times \text{ }\pi \times 21\times 21 \\
 & =\dfrac{1}{12}\times \text{ }\dfrac{22}{7}\times 21\times 21 \\
 & =\dfrac{1}{4}\times 22\times 21 \\
& =\dfrac{1}{2}\times 11\times 21 \\


 & =\dfrac{231}{2}c{{m}^{2}} \\
\end{align}\]
 \[\left[ take\text{ }\pi \text{ }=\dfrac{22}{7} \right]\]
Thus we got the area of sector AOB \[=\dfrac{231}{2}c{{m}^{2}}\].\[....... \left( 2 \right)\]
Now let us find the area of sector COD.
Here \[\theta ={{30}^{\circ }}\] and radius, r = OC =7cm
\[\therefore Area\text{ }of\sec tor\text{ }COD\text{ }=\dfrac{\theta }{360}\times \text{ }\pi {{r}^{2}}\]
\[\begin{align}
  & =\dfrac{30}{360}\times \text{ }\dfrac{22}{7}\times 7\times 7 \\
 & =\dfrac{1}{12}\times 22\times 7 \\
 & =\dfrac{11\times 7}{6} \\
 & =\dfrac{77}{6}c{{m}^{2}} \\
\end{align}\]
Thus we got area of sector COD \[=\dfrac{77}{6}c{{m}^{2}}\]\[...... \left( 3 \right)\]
Now let us substitute the value of (2) and (3) in equation (1)
\[\therefore \text{Area of}\] Shaded region = Area of sector AOB – area of sector COD
 \[\begin{align}
  & =\dfrac{231}{2}-\dfrac{77}{6} \\
 & =\dfrac{693-77}{6} \\
 & =\dfrac{616}{6} \\
 & =\dfrac{308}{3} \\
 & =102.67c{{m}^{2}} \\
\end{align}\]
Thus we got the area of the shaded region as \[102.67\text{c}{{\text{m}}^{\text{2}}}\].

Note: The sector of the circle can also be said as a pizza slice. A circular sector of a circle sector is the portion of a disk enclosed by 2 radii and an arc. Thus you need to remember the area of sector formed by a circle to solve problems of a similar kind.