Question

Question 15
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.

Answer 1

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

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

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

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

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

Area of sector with $\angle D=\frac{\angle D}{{360}^{\circ }}\times \pi r^2=\frac{\angle D}{{360}^{\circ }}\times \pi \times (21{)}^{2}c{m}^2$

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

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

$[\because sumofallinterioranglesinanyquadrilateral={360}^{\circ }]$

$=22\times 3\times 21=1386c{m}^2$

Hence, required area of the shaded region is $1386c{m}^2$