In figure, arcs have been drawn of radius 21 cm


In figure, arcs have been drawn of radius 21 cm each with vertices A, B, C and D of quadrilateral ABCD as centres. Find the area of the shaded region.


Given that, radius of each arc (r) = 21 cm

Area of sector with $\angle A=\frac{\angle A}{360^{\circ}} \times \pi r^{2}=\frac{\angle A}{360^{\circ}} \times \pi \times(21)^{2} \mathrm{~cm}^{2}$

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

Area of sector with $\angle B=\frac{\angle B}{360^{\circ}} \times \pi r^{2}=\frac{\angle B}{360^{\circ}} \times \pi \times(21)^{2} \mathrm{~cm}^{2}$


Area of sector with $\angle C=\frac{\angle C}{360^{\circ}} \times \pi r^{2}=\frac{\angle C}{360^{\circ}} \times \pi \times(21)^{2} \mathrm{~cm}^{2}$

and area of sector with $\angle D=\frac{\angle D}{360^{\circ}} \times \pi r^{2}=\frac{\angle D}{360^{\circ}} \times \pi \times(21)^{2} \mathrm{~cm}^{2}$

Therefore, sum of the areas $\left(\right.$ in $\left.\mathrm{cm}^{2}\right)$ of the four sectors

$=\frac{\angle A}{360^{\circ}} \times \pi \times(21)^{2}+\frac{\angle B}{360^{\circ}} \times \pi \times(21)^{2}+\frac{\angle C}{360^{\circ}} \times \pi \times(21)^{2}+\frac{\angle D}{360^{\circ}} \times \pi \times(21)^{2}$

$=\frac{(\angle A+\angle B+\angle C+\angle D)}{360^{\circ}} \times \pi \times(21)^{2}$

$\left[\because\right.$ sum of all interior angles in any quadrilateral $\left.=360^{\circ}\right]$

$=22 \times 3 \times 21=1386 \mathrm{~cm}^{2}$

Hence, required area of the shade region is 1386 cm²

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