Answer
$\approx 11,781$ f$t^{2}$
Work Step by Step
Starting from the bottom right, counterclockwise, there
are four quarter-circles (central angle $\displaystyle \frac{\pi}{2}$) with radii
$r_{1}=100-60=40$ ft
$r_{2}=100$ ft
$r_{3}=100-50=50$ ft
$r_{4}=100-50-20=30$ ft
The area $A$ of a sector with central angle of $\theta$ radians is
$ A=\displaystyle \frac{1}{2}r^{2}\theta$.
The total area of the 4 quarter-circles is
$ A=\displaystyle \frac{1}{2}\cdot\frac{\pi}{2}(40^{2}+100^{2}+50^{2}+30^{2})\approx$11780.972451
$\approx 11,781$ f$t^{2}$
