SE pe. d'A x 12 pi. 1 por) à R. 97 pi. 55 po. car
5. (154 x 12,8) + (26 x 148) + (1885 x 82) x 706.7260 to. car. ; "706: 7250 + 3° = 353.36376 to. car.; 303.363175 + 9 = R. 0 arp. 39.26 + to. car.
PROBLÈME V,— EXERCICES (p. 276).
1. (15 + 20 + 25) + 2 = 30, demi-somme des côtés; les trois restes sont 30 — 15 = = 15, 30 — 20 = 10, 30 — 25 = 5;
30x15 x 10x5 = y 22500 = R. 150 pi. car., ou 162 ver. A 2. (2450 + 25124 + 12843)—+ 2 — 3123.5; 3123.56 — 2460 = 673, 5, 3123.5 — 2512.5 — 611, 3123. 5 — 184.5 = 1839; 3123.5 x
673.5 x 611 x 1839 = 2363752764740.25; y ETEULPAT TUE 25 = 1537450.08 ; 1537450.08 + 100000 = R. 15.3745 + acres — 15 À. 1 R. 19.92 per.
8. (30 + 22.5 + 22.5) + 2 = 37.5; 317.5 — 30 = 7.5, 37.5 — 22.5 = ee 37.5 x 7.5 x 22.5 x 22.5 = 63281.25 ; y 63231.26 = R. 251.5
4 8 pi. ÉD OL bS 10 pi. 4 po. = 124 po., 15 pi. = 180 po. ; (101 + 124 + 180) + 2 = 202.5; 202.5 — 101 = 10.5, 202.8 — 124 = 78,5, 202.5 — 180 = 22.5; 202.5 x 101.5 x 78.5 x
22.5 := 36303060.9375 ; y 36303060. RES = 6025.20 ; 6025.20 + 1296 = R. 4.649 ver. car
5. 16 to. 3 pi. = 93 pi., 248 to. = 146 pi., 36 to. 6 pi. = 221 pi.; (93 + 146 + PNETE = 230 ; 230 — 93 = 137, 230 — 146 = 84, 230 — 221 = 9; 230 x 137 x 84 x 9 = 23821560 pi. Cur.; 23821560 = 4880.733 + pied car.; 4880.733 + 32400 = R. 0 arp. 15.063 per. car.
6. (570 + 630 + 1147.5) + 2 = 1173.75; 1173.75 — 570 603.75, 1173.75 — 630 — 543.75, 1173.76 — 1147.6 = 26.25 ; 1173. 75 x 603.75 x 543.75 x 26.25 = 1011483786. 6191 + 3 1 10114893786.6191 = 100572.828 + ver. car. = R. 20 A.3 R. 4,72 per.
PROBLÈME VI,— EXERCICES (p. 277).
1, 352 = 1225; 1225 + 4 = 306.25 ; 1225 — 306.25 — 918.75, 918.75 x 306. Les = = 281467. 1875; V281367.1875 = R. 530.44 per.
2. 14 = 03 ver; N = 1; CP QER Age + 4 = Mg jan es : | wie 1] = ARÉST; AEERT x GpAP = 1083GQhA ; 1105459328 = 10269.32; 10269.32 + 256 = R,. 40.11 ver. car,
8. 19 15 pi. 8 po. = 188 po. ; 1852 = 35344; 35344 + 4 = 8836; 35344 — É836 = 26508 ; 26508 x 8836 — 2341224688 | ; y 234224688 — R. 15304, 40 po. car. ou 106.28 + pi. car. ; 29 18 pi. 10° 8” = 2720” ; 27202 = 7398400 ; 7398400 + 4 = 1849600 ; 1398400 — 1549600 = = 5548800 ; 5848800 x 1849600 = 102630604800 ; 7 10263060480000 = R. 3203601" = 154.49 pi. car. ; 39 812 = 6561 ; 6561 + 4 = 1640.25; 6561—1640.25—4920.75 ; 4920.75 x 1640.25 = 8071260.1876 ; y 8071260.1876 = R. 2840. 996 ver. car.
PROBLÈME VIT,— £xERGICES (p. 277).
1. 1A.3R. = 280 perches; 280 - 54 = 5019; 5019 x 2 = R. 101$. perches.
2. Larp. = 1.1787 acre; 1.1787 x 7 — 8.2509 acres; 8.2509 x 4840 = 39934.356 verges; 39934356 + 4 = 9983, 589 ; 9983. 589 x 2 = R. 19967.178 verges.
PROBLÈME ÎX,— EXERCICES (p. 278).
1. 1442 = 20736; 84 ver. — 252 pi.; 2622 — 63504; 20736 + 63504 — 84240 ; 4 84240 = R. 290.24 + pieds.
2. 502 = 2500 ; 402 = 1600; 302 = 900; 2500 — 1600 = 900 ; #900 — 30 ; 2500 — 900 = 1600 ; : N1600 : æ 40; 30 + 40 = R. 70 pieds.
3. (12432 = 1564; (174)? = 3064 ; 1561 + 306$ = 4624; / 4624 = R, 214 piols,-
4, 12.62 — 166.25; 156.25 + 156.25 = 312.5; 4 312.60 = R. 17.67 pieds.
6. 17.62 = 306.25; 306.25 + 306.25 = 612.6; 4612.50 = R. 24.78 pieds.
MXSURE DES SURFACES 116
PROBLÈME X,-- EXBROICES (p. 278).
1. 1062 &: 11236 ; 902 = 8100; 11236 — 8100 = 3136 ; y 3136 = R. 56 pieds. PRET
2. 60% = 2500; 1602 = 25600 ; 25600 — 2500 = 23100; 23100= R. 161.98 pieds. FE
3. 19102 + 122 = 244 ; ÿ 244 = 15.6204 pi. ; 2° 122=144 ; 144+ 16.6204 = 9.2186 ; 9.218662 = 84,98258696 ; 144 — 84,98258596 æ= 069.01741404 ; 4 59.01741404 = 7,68 ; 7.68 + 30 + 7.68= 37.68 ; 39 15.6204— 9.2186 = 6.4018.
PROBLÈME XI,— EXEROIOES (p. 279).
1. Le quadrilatère est divisé en deux triangles, savoir: A B C, ACD; (12 + 15 + 22) + 2 = 24.5; 24,5 — 12 = 12.5, 24.5— 15 = 9,5, 24.5— 22 = 2,5; 24,5 x 12.5 x 9.6 x 2,5 = 7273. 43175 ; 11273.4375 = 85.28 pi. car., surface d’un triangle ; (10 + 18 + 22) + 2 = 255 25 — 10 = 15, 25 — 18 = 7, 25 — 22=3;25 x 15 x 7 x 3 = 7875; y 1875 = 88.74 pi. car., surface du second triangle ; 88.74 + 85.28 = R. 174.02 pi. car.
2. 12.25 + 15.06 = 27.80 ; 27.380 x (40.25 + 2) = R. 549.4125 pieds carrés.
8. 66.25 + 60.75 = 117; 117 x (108,5 + 2) = R. 6347.25 pi. car.
6. 402 — 1600, 342 = 1156; 1600 — 1156 = 444; Y 444 = 21.07, distance de À à F; 362 = 1296, 35.10% = 1232.01; 1296 — 1232.01 = 63.99; 4 63.99 = 7.99, distance de C à E; toute la diagonale est égale à 21.07 + 8 + 7.99 = 37.06; 34 + 35.10= 69.10 ; 69.10 x (37.06 + 2) = R. 1280.42 + ver. car.
6. 253% — 203 — 625 — 400 = 225; y 225 = 15, distance de À à F'; 223 — 213 = 484— 441 = 43; y 43 = 6. 65, distance de C à E; 15 + 7 + 6.565 = 28.55, ou toute la diagonale; (20 + 21) x (28.55 + 2) = R. 585.275 ver. carrées.
PROBLÈME XII,— EXERCICES (p. 280).
1. 1916 x [(24 + 36) + 2] = R. 480 pi. car.; 2° 20.15 x [ (34.25 + 62.49) + 2] = KR. 974.6555 ver. car.; 3° 36.20 x ( (75.70 + 85.80) +2] R. 2923.15 pi. car. ; 40 55,50 x [ (106.50 + 134.45) + 2 ]= R. 6686.3625 ver. car. ; 5° 70.25 x L (145.75 + 109.25) + 2]== R. 8956.8750 pi. car.
