A complete and practical solution book for the common school teacher . 5 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=
A complete and practical solution book for the common school teacher . 5 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; then KT : AO::l : 3. PROBLEM 367. From a circular piece of pasteboard whose diameter measures 60 in.,John cut the larg-est circles that could be measured from it; Mary cutthe two larg-est circles from what remained: what is the area of one ofJohns and Marys pieces? Solution.(1) From Fig. 53, let AB be the diameter of the given circle,. MENSURA TION. 177 AO the diameter of one of Johns and KT the diameterof one of Marys pieces. (2) Then AO=30 in., or S0 = 15 in. (3) The area of AO = 1527r=2257r sq. in. (4) E0=| of 60, or 20 in. Then, one of Marvs pieces^ 102tt, or IOOtt. PROBLEM 368. A circular field is inclosed by 12-foot boards: how many acres inthe field, if there are as many acres in the field as there are boards inthe fence surrounding- it, the fence being 4 boards high? Solution. (1) Let AB be the diameter of the circle. (2) AO=R, the radius. (3) (4) =the number of boards. La (5) One acre contains 43560 sq. ft. R 27T (6) •• <or^ —the number of acres in the fig. 54. 4dobO field. (7) Since the number of boards in the circumference equals t| U C • C u 8R7T R27T the number or acres in the held,
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