\fdO'^ - Old Forge Coal Mines

sq.
GEOMETRY AND TRIGONOMETRY. 161
Diameter of cross-section of ring = 1^ inches.
Area of cross-section of ring=(l-J X. 7854 = 1.76715
in. Ans.
Volume of ring = 1.76715 X 45.553 = 80.499 cu. in. Ans.
{b) Weight of ring = 80.499 X .261 = 21 lb. Ans.
be solved like the one in Art.
790. A quicker method of solution is by means of the
principle given in Art. 826.
(344) The problem may
(345) The convex area = 4 X 5j X 18 = 378 sq. in. Ans.
Area of the bases = 5— X 5- X 2 = 55.125 sq. in.
4 4.
Total area = 378 -f 55.125 = 433.125 sq. in. Ans.
Volume = (5^) X 18 = 496.125 cu. in. Ans.
(346) InFig. 16, (9C=-^^„. /^ of 360° = 60°, and
since ^ O C =^^ oi A OB, A (9(7=30°).
Area of AOB= ^^^^^'^^^ = 62.352
square feet.
Since there are 6 equal triangles in a fig. i6.
hexagon, then the area of the base = 6 X 62.352,
square feet.
Perimeter = 6 X 12,
or 72 feet.
Convex area = = 1,332 sq. ft. Ans.
or 374.112
Total area = 1,332 + 374.112 = 1,706.112 sq. ft. Ans.
(347) Area of the base = 374.112 square feet, and altitude
= 37 feet. Since the volume equals the area of the
base multiplied by — of the altitude, we have
Volume = 374.112 X ^ = 4,614 cubic feet.
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