Solution
Given AB = \[18\] cm, DC = \[32\] cm
Given, Distance between AB and DC = Height = \[14\] cm
We know that Area of the trapezium = (\[1/2\]) × (Sum of parallel sides) × Height
= \[(1/2)\times (18+32)\times 14\] = \[350\] \[c{{m}^{2}}\]
Given AB ∥ DC, ∴ \[\angle A+\angle D={{180}^{\circ }}\]and \[\angle B+\angle C={{180}^{\circ }}\]
Also given, radius of each arc = \[7\] cm
Therefore,
Area of the sector with central angle as A = \[(1/2)\times (\angle A/180)\times \pi \times {{r}^{2}}\]
Area of the sector with central angle as D = \[(1/2)\times (\angle D/180)\times \pi \times {{r}^{2}}\]
Area of the sector with central angle as B = \[(1/2)\times (\angle B/180)\times \pi \times {{r}^{2}}\]
Area of the sector with central angle as C = \[(1/2)\times (\angle C/180)\times \pi \times {{r}^{2}}\]
Hence, Total area of the sectors,
= \[\frac{\angle A}{360}\times \pi \times {{r}^{2}}+\frac{\angle D}{360}\times \pi \times {{r}^{2}}+\frac{\angle B}{360}\times \pi \times {{r}^{2}}+\frac{\angle C}{360}\times \pi \times {{r}^{2}}\]
= \[\left( \frac{\angle A+\angle D}{360}\times \pi \times {{r}^{2}} \right)+\left( \frac{\angle B+\angle C}{360}\times \pi \times {{r}^{2}} \right)\]
= \[\left( \frac{180}{360}\times \frac{22}{7}\times 49 \right)+\left( \frac{180}{360}\times \frac{22}{7}\times 49 \right)\]
=\[77+77\]
= \[154\]
∴ we will get Area of shaded region = Area of trapezium – (Total area of sectors)
= \[350-154\] = \[196\] \[c{{m}^{2}}\]
therefore, the area of shaded region is \[196\] \[c{{m}^{2}}\].