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Note. When the quotient is not found exactly in the table, proportion may be made between the next less and greater 2-ea, in the same manner as is done for logarithms, or any other table.

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Ex. 2. Taking the same example as before, in which are given the chord AB 12, and the radius 10, or diameter 20.

And having found, as above, DE = 8; then ce — DE = co = 10 – 8 = 2. Hence, by the rule, cD + cF = 2 -- 20 = 1 the tabular height. This being sought in the first column of the table, the corresponding tabular area is found = -0.4088. Then '04088 x 20° = -0.4088 x 400 = 16:352, the area, nearly the same as before.

Ex. 3. What is the area of the segment, whose height is 18, and diameter of the circle 50 ° Ans. 636-375.

Ex. 4. Required the area of the segment whose chord is 16, the diameter being 20 ° Ans. 44-728 PROBLEM xIII.

circle whose diameter is 1 ; and the first column contains the

corresponding heights or versed sines divided by the diameter.

Thus then, the area of the similar segment, taken from the table,

and multiplied by the square of the diameter, gives the area of the segment to this diameter.

PROBLEM

-**

To measure long Irregular Figures.

TAKE or measure the breadth at both ends, and at several places, at equal distances. Then add together all these intermediate breadths and half the two extremes, which sum multiply by the length, and divide by the number of parts, for the area”.

Note. If the perpendiculars or breadths be not at equal distances, compute ali the parts separately, as so many trapezoids, and add them all together for the whole area.

Or else, add all the perpendicular breadths together, and divide their sum by the number of them for the mean breadth, to multiply by the length; which will give the whole area, not far from the truth.

Ex. 1. The breadths of an irregular figure, at five equi- distant places, being 8:2, 7-4, 9:2, 10.2, 8:6; and the whole length 39; required the area 3

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* This rule is made out as follows: —Let ABCD be the irregular piece; having the several breadths AD, EF, GH, IK, BC, at the equal distances AE, EG, G 1, 1 B. Let the several breadths \in order be denoted by the corresponding letters a, b, c, d, e, and the whole length AB by l; then compute the areas of the parts into which the figure is divided by the perpendiculars, as so many trapezoids, by prob. 3, and add them all together. Thus, the sum of the parts is, a + b b -- c

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Ex. 2. The length of an irregular figure being 84, and the breadths at six equidistant places 17 4, 20-6, 1+2, 16-5, 20. 1, 24'4; what is the area Ans. 1550-64.

PROBLEM xIV.

To find the Area of an Ellipsis or Oval.

MULTIPLY the longest diameter, or axis, by the shortest; then multiply the product by the decimal '7854, for the area. As appears from cor. 2, theor. 3, of the Ellipse, in the Conic Sections.

Ex. 1. Required the area of an ellipse whose two axes

are 70 and 50. Ans. 2748-9. Ex. 2. To find the area of the oval whose two axes are 24 and 18. Ans. 339:2928.

PROBLEM XV.
To find the Arca of an Elliptic Segment.

FIND the area of a corresponding circular segment, having the same height and the same vertical axis or diameter. Then say, as the said vertical axis is to the other axis, parallel to , the segment's base, so is the area of the circular segment before found, to the area of the elliptic segment sought. This rule also comes from cor. 2, theor. 3, of the Ellipse.

Otherwire thws. Divide the height of the segment by the vertical axis of the ellipse; and find, in the table of circular segments to prob. 12, the circular segment having the above quotient for its versed sine: then multiply all together, this segment and the two axes of the ellipse, for the area.

Ex. 1. To find the area of the elliptic segment, whose height is 20, the vertical axis being 70, and the parallel axis 50. - t

which is the whole area, agreeing with the rule: m being the arithmetical mean betweet, the extremes, or half the sum of them both, and 4 the number of the parts. And the same for any other number of parts whatever.

Here

Here 20 -- 70 gives 284 the quotient or versed sine; to which in the table answers the seg. 18518 then 70

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Ex. 2. Required the area of an elliptic segment, cut off parallel to the shorter axis ; the height being 10, and the two axes 25 and 35. Ans. 162°03.

Ex. 3. To find the area of the elliptic segment, cut off 4 parallel to the longer axis; the height being 5, and the axes 25 and 35. - Ans, 97°8425.

PROBLEM XVI,

To find the Area of a Parabola, or its Segment.

Multiply the base by the perpendicular height; then take two-thirds of the product for the area. As is proved in theorem 17 of the Parabola, in the Conic Sections.

Ex. 1. To find the area of a parabola; the height being 2, and the base 12.

Here 2 x 12 = 24. Then 3 of 24 = 16, is the area.

Ex. 2. Required the area of the parabola, whose height is 10, and its base 16. Ans. 1063.

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BY the Mensuration of Solids are determined the spaces included by contiguous surfaces; and the sum of the measures of these including surfaces, is the whole surface or superficies of the body. The measure of a solid, is called its solidity, capacity, or COłłtent. Solids are measured by cubes, whose sides are inches, or feet, or yards, &c. And hence the solidity of a body is said to be so many cubic inches, feet, yards, &c, as will fill its capacity or space, or another of an equal magnitude. T - The

The least solid measure is the cubic inch, other cubes being taken from it according to the proportion in the following table, which is formed by cubing the linear proportions.

Table of Cubic or Solid Measurer.

1728 cubic inches make 1 cubic foot
27 cubic feet - 1 cubic yard
1663 cubic yards - 1 cubic pole
64000 cubic poles - 1 cubic furlong
512 cubic furlongs - 1 cubic mile.

PRO E L E M I.
To find the Superficies of a Prism or Cylinder.

MULTIPLY the perimeter of one end of the prism by the length or height of the solid, and the product will be the surface of all its sides. To which add also the area of the two ends of the prism, when required “.

Or, compute the areas of all the sides and ends separately, and add them all together.

Ex. 1. To find the surface of a cube, the length of each side being 20 feet. Ans. 2400 feet.

Ex. 2. To find the whole surface of a triangular prism, whose length is 20 feet, and each side of its end or base 18 inches. Ans. 91-948 feet.

Ex. 3. To find the convex surface of a round prism, or cylinder, whose length is 20 feet, and the diameter of its base is 2 feet. Ans. 125'664.

Ex. 4. What must be o: for lining a rectangular cistern with lead, at 3d. a pound weight, the thickness of the lead being such as to weigh 7lb. for each square foot of surface; the inside dimensions of the cistern being as follow, viz. the length 3 feet 2 inches, the breadth 2 feet 8 inches, and depth 2 feet 6 inches 2 - Ans. 31. 53. 93d.

* The truth of this will easily appear, by considering that the

sides of any prism are parallelograms, whose common length is

the same as the length of the solid, and their breadths taken all together make up the perimeter of the ends of the same.

And the rule is evidently the same for the surface of a cylinder.

- PRORLEM

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