x 0.66% x 0.7 == R. 0.689644 ver, cub

1.98 + (31416 x 2) = 0.315 ver. == le rayon; 3.1416 x 0.315 x 0.315 = 03117 ver. car. == la base; le volume = 0.3117 x (1.23 + 3) = R. 0.127797 ver. cub.

. Le rayon = (52 — 47) = 3 ver, Le voluine == (3.1416 x

32 x 4) + 3 = KR, 37.699200 ver. cub.

(1.6 x 3) + 0.80 = R. 6 ver. car.

4 + (3.60 + 3) mm R. 3.33 verges.

PROBLÈME XIII,— ExEROICES (p. 312).

[G.4 x 2) + (2.1 x 2)] + 23 = 3.60 ver.; 3.1416 x 3.5 x 2.6 = KR, 28.5885 ver. car.

. (6 x 8.48) + 2 = R. 25.44 pi. car.

. LCL x 2) + (2.8 x 2)] + 2 = 4.90 ver. ; 3.1416 x 4.9 x 3=

R. 46.181565 ver. car.

. (2.30 + 2.10) + 2 = 2.20 verges; 3.1416 x 2.20 x 3.54 =

R. 26.5402 ver. car.

42.6 + (1.95 + 2) = 48.689 ver. = La somme des circonférences des bases; 3.1416 x (1.4 x 2) = 8.796 verges = la circont.

supérieure; 43.589 — 8.796 = 34.793 ver. = la circonf. inférieure; 34.793 + 31416 = 11.074 = le diamètre inférieur ; 11.074 + 2 = R. 5.537 verges

PROBLÈME XIV,-—RXEROICES (p. 312),

. (0.90 +3) x (2.25 + 1.21 + 42:25 x L.21) = R, 1.633 v. cub,

. (84 xX 3) + (3 + 134 + (8 x 12) = R. 12 pie . 31416 x 0.422 = 0.654178 ver. car. = la base supérieure ;

3.1416 x 0.632 = 1.246901 ver. car. = la base inférieure ; 1 (0.554178 x 1.246901) = 0.831267 ver. car. = la moyenne des bases ; (2.1 + 3) x (0.654178 + 1.246901 : 0.831267) = R. 1.842642 ver, cuh.

. (A5 + 0.628) + 2 = 1.0365 ver, = la circonf. moyenne :

1.0365 + 3.1416 = 0.329 = le diamètre de la circonf. moyenne; 3.1416 x (0.329 + 4) = 0.2584 ver. car. = la sur£ de la section moyenne ; 0.2584 x 6.75 = R. 1.74420 ver. cub.

PROBLÈME XV,— EXEROIOES (p. 313).

. 3.1416 x 3.08 x 2 = lère R. 19.352 ver. ; 3.1416 x 2.08 x