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amid or Cone.

ltiply that area by the the product for the

1 square pyramid, each endicular height 25. Ans. 7500.

angular pyramid, whose side of the base 3.

Ans. 38.97117.

triangular pyramid, its e three sides of its base Ans. 71-0352.

pentagonal pyramid, its of its base 2 feet?

Ans. 27.5276.

the hexagonal pyramid, le of its base 6 inches? Ans. 1.38564 feet.

cone, its height being s base 9 feet.

Ans. 22.56093.

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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 area, in the same manner as is done for logarithms, or any other table.

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0100133 11 0470121 11990 31207384130319
02003751205339 22 12811 32 2166742 31304
030068713 06000 23 13646 33 226034332293
0401054 14 06683 24 144934 235474433284
05 0146815 07387 25 15354 35 24498||45|34278
06 01924 160811126 16226 36 254554635274
0702417 17 08853271710937264184736272
0802944 18 09613 28 180023827386 48-37270
090350219 10390 29 18905 39.2835949-38270
∙10·04088·20 •11182||·30 •19817 •40·29337||·50-39270|||

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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 CD 10 82. Hence, by the rule, CDCF = 2 20 = 1 the tabular height. This being sought in the first column of the table, the corresponding tabular area is found04088. Then 04088 x 202=04088 × 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

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.



To measure long Irregular Figures.

TAKE or measure the breadth at both ends, and at se, veral 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 equidistant places, being 8.2, 74, 92, 102, 86; and the whole length 39; required the area?

8.2 8.6

2) 16.8 sum of the extremes.

8.4 mean of the extremes.




35.2 sum.




4) 1372-8

343-2 Ans.

35.2 sum.

*This rule is made out as follows: -Let ABCD be the irregular piece; having the several breadths AD, EF, CH, IK, BC, at the equal distances AE, EG, GI, IB. Let the several breadths in order be denoted by the corre

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sponding letters a, b, c, d, e, and the whole length AB by ; 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,

d+e 2


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e × 1/1 +


= ( a + b + c + d + 1⁄2 c) × 4 l = (m + b + c + d) i l, ·

d +



11. 2. Tia rengti, of an irregur figure being 84, and the irgeditis art six equidistant pe 14, 20¤, 1# '2, 165, 201, Des: Vint & te are? Ans. 1550 64.

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Mazie ange diameter, or axis, by the shortest;

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Ix 1. Femured the area of an ellipse whose two axes are 10 m de Ans. 2748-9.

Ex 2. To End the arm of the cul whose two axes are 24 and .3 Ams. 339-9928.


To find the Area 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.

Other site thus. 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.

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

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ch is the whole area, agreeing with the rule: m being the >tical mean between the extremes, or half the sum of them of 4 the number of the parts. And the same for any other of part whatever,


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