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In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = \[18\] cm, DC = \[32\] cm and distance between AB and DC = \[14\] cm. If arcs of equal radii \[7\] cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.
In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = \[18\] cm, DC = \[32\] cm and distance between AB and DC = \[14\] cm. If arcs of equal radii \[7\] cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.
Areas Related to Circles | Exercise 11.4 | Maths | NCERT Exemplar
In Fig. 11.17, ABCD is a trapezium with AB || DC, AB = \[18\] cm, DC = \[32\] cm and distance between AB and DC = \[14\] cm. If arcs of equal radii \[7\] cm with centres A, B, C and D have been drawn, then find the area of the shaded region of the figure.

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 \[196\] \[c{{m}^{2}}\].

 

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