If a pool has to be constructed with a boundary of width 14 m consisting of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. Then, the area of the boundary is
7056 m2
We have, OC=O′D=35 m
and AC=BD=14 m
Also, AB=CD=FE=GH=120 m
Now, OA=O′B=35+14=49 m
Area of the region (boundary) = Area of rectangle ABCD + Area of rectangle EFGH + 2(area of semi-circle with radius 49 m − Area of semi-circle with radius 35 m)
=120×14+120×14+2(12×227×492−12×227×352)=[1680+1680+227(492−352)]=[3360+227(49+35)(49−35)]=3360+3696=7056 m2