A complete and practical solution book for the common school teacher . 59. (31) —77 2407 = .38828. (32) .38828 : x\\ 02059 : .02118. (33) .-. *=.39944. (34) X=° 38 (35) PB=2rcosX=+ rd. 192 FAIRCHILDS SOLUTION BOOK. PROBLEM 387. Three men own a grindstone 2 ft. 8 in. in diameter: how manyinches must each grind off to get an equal share, allowing 6 in. wastefor the aperture? (/?. //. A., p. 406, prod. 84.) Solution. (1) Let AT be separated into three equal parts, AH,HD and DT. (2) Let perpendiculars at the points of division bedrawn to intersect thecircumference con
A complete and practical solution book for the common school teacher . 59. (31) —77 2407 = .38828. (32) .38828 : x\\ 02059 : .02118. (33) .-. *=.39944. (34) X=° 38 (35) PB=2rcosX=+ rd. 192 FAIRCHILDS SOLUTION BOOK. PROBLEM 387. Three men own a grindstone 2 ft. 8 in. in diameter: how manyinches must each grind off to get an equal share, allowing 6 in. wastefor the aperture? (/?. //. A., p. 406, prod. 84.) Solution. (1) Let AT be separated into three equal parts, AH,HD and DT. (2) Let perpendiculars at the points of division bedrawn to intersect thecircumference construct-ed on AB as a diameter,and these points F andE be joined with B; theywill be radii that will de-termine rings of equalarea. (3) BT is the radius of the aperture, or 3 in., which subtracted from 16 in. leavesAT=13 in. (4) The area of the stone=1627r=2567r. (5) The area of the aperture is 327r=97r. (6) The area to be ground off is 256tt—97r = 2477r. (7) Each mans share will be ^ of 2477r=82|—82j7r=173|7r, area of the stone after the first FIG. 72. (8)(9) (10) (11)(12)(13)(14) 173|7r—82j7r=91^7r, area of the stone after the secondgrinds. Dividing the area of a circle by tt and extracting thesquare root give the radius. AB = V(256^T=1,6 in. HB = V(73i»-Hr) = in. SB = V(173i7r-Hr):= in. .-. The first man grinds off in; the sec-ond — in., and the third —3 = in. PROBLEM 388. Find the radius of the largest circle that can be drawn in a quad-rant of a circle, radius 20 in. Solution. (1) Bisect the given arc TB in S. (2) Let fall the perpendicular SF, join O with S, and produce it, making SP=SF. MENSURATION. 193 (3) Join P and F; draw SL parallel to PF, OL to OT. (4) Then with the center O and radius CS=r, de-scribe the circle SLEand it will be the in-scribed circle. (5) Since the triangles OSL and OPF are similar, and OF = OS,or20-^V2 = in.; then by simi-lar triangles, we haveOP : OS:: OF :
Size: 1636px × 1527px
Photo credit: © The Reading Room / Alamy / Afripics
License: Licensed
Model Released: No
Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
