Question

In the given figure, ABCD is a trapezium with AB || DC, AB = 18 cm DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then find the area of the shaded region.
(Use $\mathrm{\pi }=\frac{22}{7}$ [CBSE 2014]

Answer 1

Area of shaded region = Area of trapezium ABCD − Area of 4 sectors

$$\begin{gathered} =\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)\times 14-\left(\frac{\angle \mathrm{A}}{360°}\mathrm{\pi }{r}^2+\frac{\angle \mathrm{B}}{360°}\mathrm{\pi }{r}^2+\frac{\angle \mathrm{C}}{360°}\mathrm{\pi }{r}^2+\frac{\angle \mathrm{D}}{360°}\mathrm{\pi }{r}^2\right) \\ =\frac{1}{2}\left(\mathrm{AB}+\mathrm{DC}\right)\times 14-\left(\frac{\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}}{360°}\right)\mathrm{\pi }{r}^2 \\ =\frac{1}{2}\left(18+32\right)\times 14-\frac{22}{7}{\left(7\right)}^2 \\ =350-154 \\ =196{\mathrm{cm}}^2 \end{gathered}$$
=24.5−14π(AB)2=24.5−14×227(3.5)2=24.5−9.625=14.875 cm2
Hence, the area of shaded region is 196 cm²