THE I R R I G A T I X AGE. 



869 



(Continued from page 863.) 



6 Elements of Sectors and Segments. In Fig. 45 let AB 

 be any chord, draw radii AO and BO and draw OM per- 

 pendicular to AB, then AM = BM 

 and arc AC_ = arc BC. Let r 

 radius of circle and c = length of 

 chord, then the other elements can be 

 computed. Let MO = y , then y = = r 



or y = \/ r 1 -- f"/4. 



y can also be found without ex- 

 tracting the square root as follows : 

 let angle AOB = a , then angle AOM 



a c a 



Sin - = from which angle can be found; then, 

 2 /* 2 



a a 



since MO -?-; = cos , MO = r cos . 



2 9 



The area of sector AOB is found as follows: the area 

 of the whole circle = T r; the area of a sector whose 



Trr~ 

 central angle = 1 is , hence if there are a degrees in 



360 



angle AOB the area of sector A OB = 



360 

 The area of segment ABC is found by subtracting tri- 



7r'"a 

 angle AOB from sector AOB, hence segment ABC = - 



c a 360 



r cos . 

 2 2 



a 



The middle ordinute CM = r r cos --, or bv simplify- 

 a 2 



ing: CM = ; (1 cos ). 



2 



Applied Problems. In a circle having a radius of 7' 9" 

 is drawn a chord 10 ft. long; find central angle, area of 

 sector, area of segment and middle ordinate. 



Referring to Fig. 45, call central angle a . then sin 



2 



- 5 -i- 7.75 =i .64052 ; refer to a table of natural sines, 



a 

 this gives = 39 50', hence angle a = 79 40'. 



MO = r cos =: 7.75 X .76791 = 5.95131 ft. 

 2 



Then middle ordinate = 7.75 5.95131 = 1.78969 ft. 



Area of sector = 3.1416 X 7.75 X 7.75 X 79.67 -4- 360 

 = 41.765 sq. ft. 



c a 



Area of triangle AOB = r cos =; 5 X 5.95131 

 = 29.76655 sq. ft. 2 2 



Area of segment ACB = 41.765 29.767 = 11.998 sq. ft. 



Problem. If in the preceding problem the angle a a nd 

 radius r are given, then the chord, middle ordinate, arc, 

 area of sector, area of segment can be computed. Let a 

 = 52 and r = 200 ft., then angle AOM = 26 and let 



c c 



c = chord ; then --- - 200 = sin 26 or = 200 X .43837 



2 2 



= 87.674 ft., hence r = 175.348 ft. 



MO = 200 cos 26 = 200 X .89879 = 179.758. 



Middle ordinate = 200 179.758 20.242 ft.. 



The length of arc = 



= 181.508 ft. 



360 

 The area of sector = 



360 



180 

 3.1416 X 200 X 200 X 52 



360 



18151.47 sq. ft. 



The area of triangle ABO = 179.758 X 87.674 = 15760.10. 

 Area of segment 18151.47 15760.10 = 2391.37 sq. ft. 



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