Question

Question 13
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 $\angle P=\frac{\angle P}{{360}^{\circ }}\times \pi r^2$

$=\frac{\angle P}{{360}^{\circ }}\times \pi \times (14{)}^{2}c{m}^2$

[$\therefore$ area of any sector with central angle $\theta$ and radius $r=\frac{\pi r^2}{{360}^{\circ }}\times \theta$]

Area of the sector with central angle $=\frac{\angle Q}{{360}^{\circ }}\times \pi r^2=\frac{\angle R}{{360}^{\circ }}\times \pi \times (14{)}^{2}c{m}^2$

Therefore, sum of the areas ( in $c{m}^2$) of three sectors

$=\frac{\angle P}{{360}^{\circ }}\times \pi \times (14{)}^2+\frac{\theta }{{360}^{\circ }}\times \pi \times (14{)}^2+\frac{\angle R}{{360}^{\circ }}\times \pi \times (14{)}^2$

$=\frac{\angle P+\angle Q+\angle R}{360}\times 196\times \pi =\frac{{180}^{\circ }}{{360}^{\circ }}\times 196\pi c{m}^2$

[ Since sum of all interior anlges in any triangles in any triangle is ${180}^{\circ }$ ]

$=98\times \pi c{m}^2=98\times \frac{22}{7}$

$=14\times 22=308c{m}^2$

Hence, the required area of the shaded region is $308c{m}^2$.