We then have Area ABCD=ABC+BCD, ab cb a+c .b. 2 Any Figure bounded by Straight Lines.—The method just given can evidently be extended, and we thus obtain RULE.- To obtain the area of any rectilinear figure. Divide the figure into triangles, or parallelograms, and find the area of each separately. Finally add all the results together. We will apply this rule to the case of a trapezoid or quadrilateral; this is a four-sided figure having no parallel sides. Quadrilateral, however, includes all cases, parallel or not. ABCD is the given quadrilateral. Divide into triangles as shown. Area ABCD=ABD+BCD, A h B C Fig. 4. Then = 1. BD × h1+ BD × h2, = 1⁄2 BD (h1+h2). More complex cases are solved in an exactly similar manner, the only difficulty being in deciding on the manner of dividing the area. For this no rules can be given; practice alone will enable the student to decide on the quickest method for any given case. We come now to areas enclosed by curved lines, of which the simplest is Fig. 5. The Circle. RULE. The area of a circle is obtained by squaring the diameter, and multiplying the result by .7854. Or squaring the radius and multiplying by 3.1416. The number 3.1416, of which .7854 is one-fourth, is denoted by ; and hence in the figure [For most practical purposes π may be taken as 22.] In rectilinear figures the periphery or circumference can be directly measured; but this cannot be done in curved figures, hence we require a rule. RULE. The circumference of a circle is obtained by multiplying the diameter by π. Hence in figure 5 Circumference = πd or 2πг. The Ring.—This may represent the cross-section of a hollow shaft, or column. If a table of squares be available, use (1), otherwise use (2). It is immaterial whether the hole be concentric with the outside or not. The equations are the same. But the B result in words would not be true unless by C A Fig. 7. "thickness" mean thickness be understood. Sector of a Circle.-ABCD is the sector. The angle BAD is its angle. D RULE. The area of a sector is obtained by multiplying the area of the whole circle by the ratio of the angle of the sector to four right angles. Segment of a circle.—BCD (Fig. 7) is the segment. Its area is best found by subtracting the triangle ABD from the sector ABCD. The Ellipse.-This is the figure produced when a circular cylinder is cut across in any direction not perpendicular to its axis. It is not of great importance to us, so we shall not examine its properties. There are several methods of constructing an ellipse, of which we will give one, which is simple, and assists the student to remember the value of the area. To construct an ellipse with given axes. AB and CD are the given axes, O the centre. The lengths of the semi-axes are a and b respectively, a being the major or greater. On AB, with centre O, describe a circle. Divide AB into a number of parts, and draw ordinates of the circle as shown. Reduce each ordinate in the pro portion of b to a. Then D B through the tops of the reduced ordinates draw a curve, which is the required ellipse. The method of reduction is shown for the ordinate cd. The inner circle with radius is described, d is joined to O cutting the inner circle in e, and ef is drawn parallel to AB, cutting cd in f, then ƒ is one of the points. For cf:cd::b: a (similar triangles). Only one quarter need be drawn in this manner, the others are quite symmetrical. Now each ordinate of the circle is diminished in the ratio b/a, hence evidently so is the whole area. Hence one rule will suffice both for ellipse and circle as follows. RULE. The area of an ellipse is π/4 times that of the circumscribing rectangle. The circle is then a particular case, where b=a, and the rectangle becomes a square. The Parabola.-This curve is the most important we have to deal with, its construction enabling us to solve graphically many problems. To construct a parabola of given height on a given base. First case. Let AB be the given base; and D the apex be directly over C, the middle point of AB. The parabola will then be symmetrical, CD being the given height. Complete the rectangle AEFB. Divide DE into any number of equal parts (in the figure we take five), and EA into the same number. Number the divisions 1', 2', etc., from D, and 1, 2, etc., from E as shown. Join D to 1, 2, 3, and 4. Through 1′ draw a vertical cutting Dî in I; through 2' a vertical cutting D2 in II; and so on, obtaining the points I, II, III, IV. Then the curve is drawn through A, the above four points, and D. Similarly for the other side. DC is called the axis, D the apex; and the nature of the curve is such that the distance of any point in it from DE varies as the square of its distance from DC. It will be seen that the construction fulfils this condition. For example, IV is twice as far from DC as II is, but is 4, i.e. 22, times as far from DE. The parabola may also be required to have its greatest height at some point not directly over C. AB is again the base, and D the given highest point, at a height h say over M. Join CD, and complete the parallelogram AF. The construction now proceeds exactly as in the preceding case, the ordinates being drawn parallel to CD. The two sides are now unsymmetrical, CD is not the axis, nor is D the apex. CD is, however, an axis. The curve has an important property, which we shall require to use hereafter, viz. The tangent at any point bisects the abscissa of that point. Thus take for example IV, in either figure, then the tangent at IV passes through 2', i.e. bisects D4'. Another method of constructing the curve, showing yet another property, is illustrated in Fig. 11. |