# Ex.12.3 Q14 Areas Related to Circles Solution - NCERT Maths Class 10

## Question

\({AB}\) and \({CD}\) are respectively arcs of two concentric circles of radii \(\text{21 cm}\) and \(\text{7 cm}\) and center \({O}\) (see Figure). If \(\angle {AOB} = 30^\circ \), find the area of the shaded region**.**

## Text Solution

**What is known?**

\({AB}\) and \({CD}\) are arcs of two concentric circle of radii \(\text{21 cm}\) and \(\text{7 cm}\) respectively and centre \({O.}\)

\(\angle {{ AOB }}= \,30^\circ \)

**What is unknown?**

Area of the shaded region.

**Reasoning:**

Area of the shaded region \(= \) Area of sector \({ABO}\, - \) Area of sector \({CDO}\)

Areas of sectors \({ABO}\) and \({CDO}\) can be found by using the formula of

Area of sector of angle \(\begin{align} \theta = \frac{\theta }{{{{360}^\circ }}} \times \pi {r^2}\end{align}\)

where \({r}\) is radius of the circle and angle.

with degree measure \(\theta\)

with \(\theta = {30^ \circ }\) for the both the sector \( {ABO}\) and \( {CDO}\) angle \(\theta = {30^\circ}\) and radii 21cm and 7cm respectively.

**Steps:**

Radius of the sector ABO,\( \rm{}R = OB = 21\rm{}cm\)

Radius of the sector CDO,\(\rm{}r = OD = 7\rm{}cm\)

For both the sector ABOand CDO angle,\( \theta = {90^\circ}\)

Area of shaded region \(= \) Area of sector \({ABO}\; – \) Area of sector \({CDO}\)

\[\begin{align}&= \frac{\rm{\theta }}{{360}^{\rm{o}}} \times \pi {R^2} - \frac{\rm{\theta}}{{360}^\circ } \times \pi {r^2}\\&= \frac{{\rm{\theta }}}{{{{360}^{\rm{o}}}}} \times \pi \left( {{R^2} - {r^2}} \right)\\

&= \frac{{{{30}^{\rm{o}}}}}{{{{360}^{\rm{o}}}}} \times \frac{{22}}{7}\left( {{{\left( {21cm} \right)}^2} - {{\left( {7cm} \right)}^2}} \right)\\&= \frac{1}{{12}} \times \frac{{22}}{7} \times \left( {441c{m^2} - 49c{m^2}} \right)\\&= \frac{{11}}{{42}} \times 392c{m^2}\\&= \frac{{308}}{3}c{m^2}\end{align}\]