Question

In figure, arcs have been drawn with radii 14 cm each and with centres P, Q and R. Find the area of the shaded region.

Answer 1

Given that, radii of each arc (r) = 14 cm

Now, area of the sector with central angle P

= ${\displaystyle \frac{\angle \text{P}}{{360}^{\circ }}\times \pi {\text{r}}^2}$

= ${\displaystyle \frac{\angle \text{P}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2{\text{cm}}^2}$  

Area of the sector with central angle Q

= ${\displaystyle \frac{\angle \text{Q}}{{360}^{\circ }}\times \pi {\text{r}}^2}$

= ${\displaystyle \frac{\angle \text{Q}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2{\text{cm}}^2}$

And area of the sector with central angle R

= ${\displaystyle \frac{\angle \text{R}}{{360}^{\circ }}\times \pi {\text{r}}^2}$

= ${\displaystyle \frac{\angle \text{R}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2{\text{cm}}^2}$

Therefore, sum of the areas of three sectors

= ${\displaystyle \frac{\angle \text{P}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2+\frac{\angle \text{Q}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2+\frac{\angle \text{R}}{{360}^{\circ }}\times \pi \times {\left(14\right)}^2}$

= ${\displaystyle \frac{\pi }{{360}^{\circ }}\times {\left(14\right)}^2\times [\angle \text{P}+\angle \text{Q}+\angle \text{R}]}$

= ${\displaystyle \frac{\pi }{{360}^{\circ }}\times 196\times {180}^{\circ }}$ ...[Since, sum of all interior angles in any triangle is 180°]

= 98π

= ${\displaystyle 98\times \frac{22}{7}}$

= 14 × 22

= 308

Hence, the required area of the shaded region is 308 cm².