Page:EB1911 - Volume 18.djvu/156

 The two sources of error mentioned under (a) and (b) above are closely related. Suppose, for instance, that we require the area of a circular grass-plot of measured diameter. As a matter of fact, no grass-plot is truly circular; and it might be found that if the breadth in various directions were measured more accurately the want of circularity would reveal itself. Thus the inaccuracy in taking the measured diameter as the datum is practically of the same order as the inaccuracy in taking the grass-plot to be circular.

(iii) In dealing with cases where actual measurements are involved, the error (i) due to inaccuracy of the formula will often be negligible in comparison with the error (ii) due to inaccuracy of the data. For this reason, formulae which will only give approximate results are usually classed together as rules, whether the inaccuracy lies (as in the case of Huygens’s rule) in the formula itself, or (as in the case of Simpson’s rule) in its application to the data.

21. It is necessary, in applying formulae to specific cases, not only, on the one hand, to remember that the measurements are only approximate, but also, on the other hand, to give to any ratio such as a value which is at least more accurate than the measurements. Suppose, for instance, that in the example given in § 20 the diameter as measured is 15 ft. 3 in. If we take ＝3·14 and find the area to be 26288·865 sq. in.＝182 sq. ft. 80·865 sq. in., we make two separate mistakes. The main mistake is in giving the result as true to a small fraction of a square inch; but, if this degree of accuracy had been possible, it would have been wrong to give a value which is in error by more than 1 in 2000.

Calculations involving feet and inches are sometimes performed by means of duodecimal arithmetic; i.e., in effect, the tables of square measure and of cubic measure are amplified by the insertion of intermediate units. For square, measure—

while for cubic measure—

When an area has been calculated in terms of square feet, primes and square inches, the primes and square inches have to be reduced to square inches; and similarly with the calculation of volumes. The value of for duodecimal arithmetic is 3+1/12+8/122+ 4/123+8/124+ ; so that, marking off duodecimal fractions by commas, the area in the above case is of 3, 1, 8, 4, 8×15, 3×15, 3 sq. ft.＝182, 7, 10 sq. ft.＝182 sq. ft. 94 sq. in. (or 182 sq. ft. approximately).

22. Areas of Plane Rectilinear Figures.—The following are expressions for the areas of some simple figures; the expressions in (i) and (ii) are obtained arithmetically, while those in (iii)–(v) involve dissection and rearrangement.

(i) Square: side a. Area＝a2.

(ii) Rectangle: sides a and b. Area＝ab.

(iii) Right-angled triangle: sides a and b, enclosing the right angle. Area＝ab.

(iv) Parallelogram: two opposite sides a and a, distance between them h. Area＝ha.

(v) Triangle: one side a, distant h from the opposite angle. Area＝ha.

If the data for any of these figures are other than those given above, trigonometrical ratios will usually be involved. If, for instance, the data for the triangle are sides a and b, enclosing an angle C, the area is ab sin C.

23. The figures considered in § 22 are particular cases of the trapezium, which is a quadrilateral with two parallel sides. If these sides are a and b, at distance h from one another, the area is h.(a+b). In the case of the triangle, for instance, b is zero, so that the area is ha.

The trapezium is also sometimes called a “trapezoid,” but it will be convenient to reserve this term for a different figure (§ 24).

The most important form of trapezium is that in which one of the two remaining sides of the figure is at right angles to the two parallel sides. The trapezium is then a right trapezium; the two parallel sides are called the sides, the side at right angles to them the base, and the fourth side the top.

By producing the two parallel sides of any trapezium (e.g. a parallelogram ), and drawing a line at right angles to them, outside the figure, we see that it may be treated as the difference of two right trapezia. It is, however, more simple to convert it into a single right trapezium. Let CABD (fig. 1) be a trapezium, the sides CA and DB being parallel. Draw any straight line at right angles to CA and DB (produced if necessary), meeting them in M and  Along CA and DB, on the same side of MN, take MA′＝CA, NB′＝DB; and join A′B′. Then MA′B′N is a right trapezium, whose area is equal to that of CABD; and it is related to the latter in such a way that, if any two lines parallel to AC and BD meet AB, CD, MN, A′B′, in E, G, P, E′, and F, H, Q, F′, respectively, the area of the piece PE′F′Q of the right trapezium is equal to the, area of the piece GEFH of the original trapezium. The right trapezium so constructed may be called the equivalent right trapezium. In the case of a parallelogram, the equivalent right trapezium is a rectangle; in the case of a triangle, it is a right-angled triangle.

24. If we take a series of right trapezia, such that one side (§ 23) of the first is equal to one side of the second, the other side of the second is equal to one side of the third, and so on, and place them with their bases in a straight line and their equal sides adjoining each other, we get a figure such as MABCDEFS (fig. 2), which has two parallel sides MA and SF, a base MS at right angles to these, and the remainder of its boundary from A to F rectilinear, no part of the figure being outside the space between MA (produced) and SF (produced). A figure of this kind will be called a trapezoid.

(i) If from the other angular points B, C, D, E, perpendiculars BN, CP, DQ, ER, are drawn to the base MS (fig. 2), the area is MN.(MA+NB)+NP.(NB+ PC)++RS.(RE+SF)＝ (MN. MA + MP. NB + NQ. PC++ RS. SF). The lines MA, NB, PC, are called the ordinates of the points A, B, C, from the base MS, and the portions MN, NP, PQ, of the base are the projections of the sides AB, BC, CD, on the base. (ii) A special case is that in which A coincides with M, and F with S. The figure then stands on a base MS, the remainder of its boundary being a broken line from M to S. The formula then becomes

area＝(MP . NB+NQ . PC+ . . . +QS . RE),

i.e. the area is half the sum of the products obtained by multiplying each ordinate by the distance between the two adjacent ordinates. It would be possible to regard this form of the figure as the general one; the figure considered in (i) would then represent the special case in which the two end-pieces of the broken line are at right angles to the base.

(iii) Another special case is that in which the distances MN, NP, PQ, RS are all equal. If this distance is h, then

area＝h(MA+NB+PC+. . .+SF).

25. To find the area of any rectilinear figure, various methods are available.

(i) The figure may be divided into triangles. The quadrilateral, for instance, consists of two triangles, and its area is the product of half the length of one diagonal by the sum of the perpendiculars drawn to this diagonal from the other two angular points.

For figures of more than four sides this method is not usually convenient, except for such special cases as that of a regular polygon, which can be divided into triangles by radii drawn from its centre.

(ii) Suppose that two angular points, A and E, are joined (fig. 3) so as to form a diagonal AE, and that the whole of the figure lies between lines through A and E at right angles to AE. Then the figure is (usually) the sum of two trapezoids on base AE, and its area can be G calculated as in § 24. If BN, CP, DQ, FS, GT are the perpendiculars to AE from the angular points, the ordinates NB, PC, are called the offsets from the diagonal to the angular points.

The area of the polygon, in fig. 3 is given by the expression

(AP . NB+NQ . PC+PE . QD+ET . SF+SA . TG).

It should be noticed (a) that AP, NQ,. . . . SA are taken in the cyclical order of the points ABC. . . GA, and, (b) that in fig. 3, if AN and NB are regarded as positive, then SF, TG, ET and SA are negative, but the products ET. SF and SA. TG are positive. Negative products will arise if in moving from A to E along the perimeter of either side of the figure the projection of the moving point does not always move in the direction AE.

(iii) Take any straight line intersecting or not intersecting the figure, and draw perpendiculars Aa, Bb, Cc, Dd,. . . Ff, Gg to this line. Then, with proper attention to signs,

area＝(gb . aA+ac . bB+bd . cC+. . . +fa . gG).

(iv) The figure may be replaced by an equivalent trapezoid, on the system explained in § 23. Take any base X′X, and draw lines at right angles to this base through all the angular points of the figure.