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Thus, suppose the area of the segment A C B e is required, and that the length of the arc A C B equal 93 feet, F A and FB each equal 7 feet, and the chord A B equal 8 feet 4 inches, also the perpendicular e F equal 33 feet.

9.75 × 7


=34.125 feet, the area of the sector.

8.333 x 3.75


=15.624 feet, area of the triangle.

And 34.125-15-624-18.501 feet, the area of the segment.

To find the area of the space contained between two concentric circles.

Rule.-Multiply the sum of the inside and outside diameters by their difference, and by 7854; the product is the area.

1. Suppose the external circle A B equal 32 inches, and internal circle C D equal 28 inches; required the area of the space contained between them.

32+28=60, and 32-28=4; hence 60 × 4 × 7854-188.496 in. the area.




2. The exterior diameter of the fly-wheel of a steam engine is 20 feet, and the interior diameter 18 feet; required the area of the surface or rim of the wheel.

20+18.5=38·5, and 20—18.5=1.5; hence 38.5 x 1.5 x '7854 =45.35, &c. feet, the area.

To find the area of an ellipse or oval.

Rule.-Multiply the longest diameter by the shortest, and the product by 7854; the result is the


An oval is 25 inches by 16·5; what are its superficial contents?

25 × 16.5=412·5 × ·7854=323-9775 inches, the area. Note.-Multiply half the sum of the two diameters by 3-1416, and the product is the circumference of the oval or ellipse.

To find the area of a parabola, or its segment.

Rule.-Multiply the base by the perpendicular height, and two-thirds of the product is the area. What is the area of a parabola whose base is 20 feet and height 12?

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Note. Although the whole of the preceding practical applications or examples are given in measures of feet or inches, these being considered as the most generally familiar, yet the rules are equally applicable to any other unit of measurement whatever, as yards, chains, acres, &c. &c. &c.


To find the solidity or capacity of any figure in the cubical form.

Rule.-Multiply the length of any one side by its breadth and by the depth or distance to its opposite side; the product is the solidity or capacity, in equal terms of measurement.

Application of the rule to practical purposes.

1. Required the number of cubic inches in a piece of timber 23 inches long, 7 inches broad, and 33 inches in thickness.

23.5 x 7.75 x 3.625=660·203 cubic inches.

2. A rectangular cistern is in length 8 feet, in breadth 5 feet, and in depth 4 feet; required its

capacity in cubic feet, also its capacity in British imperial gallons.

8.5 × 5.25 x 4=178.5 cubic feet, and 178.5 × 6.232 (see Table of Decimal Approximations, p. 30) =1112-412 gallons.

3. A rectangular cistern capable of containing 520 imperial gallons is to be 74 feet in length, and 43 feet in width; it is required to ascertain the necessary depth.

7.25 x 4.5 x 6.232=203.318, and 2 feet 6 inches nearly.


=2.557 feet, or

4. A rectangular piece of cast iron, 20 inches long and 6 inches broad, is to be formed of sufficient dimensions to weigh 150fbs.; what will be the depth required? 20 × 6 × 263 (see Table of Decimal Approximations, Cast


Iron, p. 30)=31·96, and; =4.69 in., or 4 and in., the thickness required. 31.96

To find the convex surface, and solidity or capacity of a cylinder.

Rule. 1.-Multiply the circumference of the cylinder by its length or height; the product is the convex surface.

Rule 2.-Multiply the area of the diameter by the length or height; and the product is the cylinder's solidity or capacity, as may be required.

Application of the rules.

1. The circumference of a cylinder is 37 inches, and its length 544 inches; required the convex surface in square feet.

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54.75 x 37.5 × 007 (see Table of Approximations) =14.371 square feet.

2. A cylindrical piece of timber is 9 inches dia

meter, and 3 feet 4 inches in length; required its solidity in cubic inches, and also in cubic feet.

3 feet 4 inches=40 inches, and 92 x 7854 x 40=2544.696 cubic inches; then 2544 696 x 00058=14759 cubic feet.

3. Suppose a well to be 4 feet 9 inches diameter, and 16 feet from the bottom to the surface of the water; how many imperial gallons are therein contained?

4.75 x 16.5 x 4.895-1822-162 gallons.

4. Again, suppose the well's diameter the same, and its entire depth 35 feet; required the quantity in cubic yards of material excavated in its formation. 4.752 x 35 x 02909 22.973 cubic yards.

5. I have a cylindrical cistern capable of holding 7068 gallons, and its depth is 10 feet; now I want to replace it with one of an equal depth, but capable of holding 12,500 gallons; what must be its diameter?


4.895 × 10-48.95, and = 255.3 15.9687 feet, or 15 feet 11 inches. 48.95

6. A cylindrical piece of lead is required 74 inches diameter, and 168fbs. in weight; what must be its length in inches?

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To find the length of a cylindrical helix, or spiral, wound round a cylinder.

Rule.-Multiply the circumference of the base by the number of revolutions of the spiral, and to the square of the product add the square of the height; the square root of the sum is the length of the spiral.

Application of the rule.

1. Required the length of the thread or screw twisting round a cylinder 22 inches in circumference 3 times, and extending along the axis 16 inches.

22 × 3·5=772=5929, and 162=256; then /5929 +256=78.64 inches.

2. The well of a winding staircase is 5 feet diameter, and height to the top landing 25 feet, the hand-rail is to make 2 revolutions; required its length.

5 feet diameter 15.7 feet circumference.

15·7 × 2·5=39-252=1540·5625, and 252-625; then
1540+625=46.5 feet, the length required.

To find the convex surface, solidity, or capacity of a cone or pyramid.

Rule 1.-Multiply the circumference of the base by the slant height, and half the product is the slant surface.

Rule 2.-Multiply the area of the base by the perpendicular height, and one-third of the product is the solidity or capacity, as may be required.

Application of the rules.

1. Required the area in square inches of the slant surface of a cone whose slant height equals 183 inches, and diameter at the base 61 inches.

6.25 × 3·1416=19.635, circumference of the base; and 19.635 x 18.75


184.078125 square inches.

2. Required the quantity of lead, in square feet, sufficient to cover the slant surface of a hexagonal pyramid whose slant height is 42 feet, and the breadth of each side at the base 4 feet 9 inches.

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