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$$ABCD$$ is a trapezium with $$AD=DC=CD=10$$ units and $$AB=22$$ units. Semi circles are drawn as shown in the figure. The area of the region bounded by these semi circles in square units is
The trapezium has height $$8$$ and area $$\frac{1}{2}(10+22)\cdot8=128$$ square units. From the construction, the net contribution of the semicircular arcs is the difference of the semicircles with diameters $$22$$ and $$10$$, namely $$\frac{\pi}{8}(22^2-10^2)=48\pi$$. Hence the required area is $$128+48\pi$$ square units.
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