Question 1182637
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Here is a sketch....<br>
{{{drawing(800,400,0,100,0,60
,line(10,10,90,10),line(90,10,90,50),line(90,50,10,50),line(10,50,10,10)
,line(10,30,25,10),line(10,30,25,50),line(10,30,30,30)
,circle(0,30,30)
,locate(10,53,A),locate(10,9,B),locate(90,9,C),locate(90,53,D)
,locate(8,30,E),locate(25,9,F),locate(25,53,G),locate(31,30,H)
,locate(7,40,20),locate(18,40,30)
)}}}<br>
Rectangle ABCD has side lengths 40 and 80; the circular arc is centered at E, the midpoint of AB. AE=20 and EG=30; the Pythagorean Theorem gives us AG=10*sqrt(5).<br>
The area of the smaller portion of the lot is the area of the two congruent triangular regions EAG and EBF, plus the area of the circle sector EFG.<br>
The areas of the triangles are easy to calculate.<br>
For the area of the circle sector, you need to find what fraction of a circle the sector is, so you need to find the measure of angle FEG.  Angle FEG is twice angle AGE; angle AGE is arcsin(20/30).  Then the area of the sector is that fraction, times the area of the whole circle, which is (pi)r^2 = 900pi.<br>
I leave it to you to do the calculations....<br>
To check your calculations, here are my figures, slightly rounded:
Angle FEG: 83.62 degrees
Area of circle sector: 656.76 m^2
Area of smaller portion of lot (circle sector plus two triangles): 1103.97 m^2
Area of larger portion: 3200-1103.97 = 2096.03 m^2
Ratio of smaller to larger: 1104:2096<br>