A complete and practical solution book for the common school teacher . ngle, and O the posi-tion of the teacher,A, B, C the positionsof the boys, and S, D,K that of the girls. (2) Let AB=c, CB=a and CA=£, and let S= i(«+*-N), (3) The area of ABC — l/s(s— a) (S—3) (S—C) FIG. 52. = 27720 sq. ft. The perimeter of the triangle:=tf-f-£-f-<::=770 ft. (4) Dividing twice the area of the triangle by the perimeter gives the radius of the inscribed circle, or the distancethe girls are from the teacher. (5) .. (2770x2)^-770 = 72 ft., the radius of the inscribed circle. (6) CA = 231 ft., AB = 250 ft. an
A complete and practical solution book for the common school teacher . ngle, and O the posi-tion of the teacher,A, B, C the positionsof the boys, and S, D,K that of the girls. (2) Let AB=c, CB=a and CA=£, and let S= i(«+*-N), (3) The area of ABC — l/s(s— a) (S—3) (S—C) FIG. 52. = 27720 sq. ft. The perimeter of the triangle:=tf-f-£-f-<::=770 ft. (4) Dividing twice the area of the triangle by the perimeter gives the radius of the inscribed circle, or the distancethe girls are from the teacher. (5) .. (2770x2)^-770 = 72 ft., the radius of the inscribed circle. (6) CA = 231 ft., AB = 250 ft. and CB = 289 ft. (7) Let *=CS and CK. Then AC—*=SA and CB—*=KB. (8) Now, AS, or AC—.*=AD, and KC, or CB—*=DB. (9) We have 231 ft.—*+289 ft.—*=250 ft., or ;r=135 ft. (10) AS = 231 ft.—135 ft. =96 ft., and 289 ft.—135 ft. (11) CO =y72MI135^ = 153 ft. AO = V72»+96a = 120 ft., and OB = v^72^+1702=170ft. Note.—This problem was first published by J. S. Royer, the pub-lisher of the School Visitor, and was solved by the author and 176 FAIRCHIL&S SOLUTION BOOK. SCALENE TRIANGLES WHOSE AREAS ARE INTEGRAL. 4 13 15 13 14 15 7 15 20 11 13 2010 17 21 12 17 25 13 20 2117 25 2617 25 2813 37 40 13 40 45 15 34 35 15 37 44 17 39 44 25 29 36 25 39 40 29 35 48 39 41 50 13 68 75 15 41 52 17 55 60 20 37 51 25 39 56 25 52 63 25 51 52 25 74 77 26 51 5529 52 69 34 65 93 35 53 66 36 61 65 37 91 66 50 39 41 39 85 92 40 51 77 41 51 5841 84 8548 85 91 50 69 73 51 52 53 52 73 7543 61 68 IV. CIRCLE. PROBLEM that AB is 3 times ET, or KT is \ of AO. Proof. (1) Let KT=r, SO=R, and ET (2) SOK is a right-angled tri- angle. (3) Then SK=R+r, hypothe- nuse. (4) OK=2R—r% the perpendic- ular. (5) SO = R, the base of the right triangle. (6) R> + (2R—r)*=(R+r)\ (7) r=|R, and R=|r. PlG. 53. (8) Now as x=2r, then *=fR. Suspose SO, or R=3, then x—\ of 3, or 4. (9) Then as SO can be applied on AB 4 times, AB = 12. (10) Then ET: AB::1:3. (11) Also, KT=2, and AO=6
Size: 1902px × 1314px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
