A complete and practical solution book for the common school teacher . 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 i
A complete and practical solution book for the common school teacher . 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 : OL, : 20 :: : OL= in. /. r= FIG.
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Keywords: ., bookcentury1800, bookdecade1890, booksubject, booksubjectgeometry
