But, inasmuch as all lines are not commensurable, we have in Geometry to treat of magnitudes and not of measures: that is, when we use the symbol A to represent a line (as in 1. 22), A stands for the line itself and not, as in Algebra, for the number of units of length contained by the line. The method, adopted by Euclid in Book II. to explain the relations between the rectangles contained by certain lines, is more exact than any method founded upon Algebraical principles can be; because his method applies not merely to the case in which the sides of a rectangle are commensurable, but also to the case in which they are incommensurable. The student is now in a position to understand the practical application of the theory of Equivalence of Areas, of which the foundation is the 35th Proposition of Book I. We shall give a few examples of the use made of this theory in Mensuration. ·Area of a Parallelogram. The area of a parallelogram ABCD is equal to the area of the rectangle ABEF on the same base AB and between the same parallels AB, FC. Now BE is the altitude of the parallelogram ABCD if AB be taken as the base. Hence area of ☐ ABCD=rect. AB, BE. If then the measure of the base be denoted by b, and ......... altitude h, the measure of the area of the will be denoted by bň That is, when the base and altitude are commensurable, measure of area measure of base into measure of altitude. S. E. Area of a Triangle. If from one of the angular points A of a triangle ABC, a perpendicular AD be drawn to BC, Fig. 1, or to BC produced, Fig. 2, B D B D and if, in both cases, a parallelogram ABCE be completed of which AB, BC are adjacent sides, area of ▲ ABC= half of area of ABCE. ` Now if the measure of BC be b, Draw the diagonals AC and BD, cutting one another in O. B D It is easy to prove that AC and BD bisect each other at right angles. Then if the measure of AC be x, measure of area of rhombus twice measure of ▲ ACD. = Area of a Trapezium. Let ABCD be the given trapezium, having the sides AB, CD parallel. Draw AE at right angles to CD. D E B Produce DC to F, making CF=AB. Join AF, cutting BC in 0. Then in As AOB, COF, ::: ▲ BA0= 4 CFO, and ▲ AOB= ▲ FOC, and AB=CF; .. A COFA AOB. Hence trapezium ABCD= ▲ ADF. I. 26. Now suppose the measures of AB, CD, AE to be m, n, p respectively; .. measure of DF=m+n, ·· CF=AB. Then measure of area of trapezium = (measure of DF × measure of AE) That is, the measure of the area of a trapezium is found by multiplying half the measure of the sum of the parallel sides by the measure of the perpendicular distance between the parallel sides. Area of an Irregular Polygon. There are three methods of finding the area of an irregular polygon, which we shall here briefly notice. I. The polygon may be divided into triangles, and the area of each of these triangles be found separately. B A E Thus the area of the irregular polygon ABCDE is equal to the sum of the areas of the triangles ABE, EBD, DBC. II. The polygon may be converted into a single triangle of equal area. If ABCDE be a pentagon, we can convert it into an equivalent quadrilateral by the following process: B Join BD and draw CF parallel to BD, meeting ED produced in F, and join BF. Then will quadrilateral ABFE=pentagon ABCDE. For ▲ BDF ▲ BCD, on same base BD and between same parallels. If, then, from the pentagon we remove ▲ BCD, and add ▲ BDF to the remainder, we obtain a quadrilateral ABFE equivalent to the pentagon ABCDE. The quadrilateral may then, by a similar process, be converted into an equivalent triangle, and thus a polygon of any number of sides may be gradually converted into an equivalent triangle. The area of this triangle may then be found. III. The third method is chiefly employed in practice by Surveyors. B M N D Let ABCDEFG be an irregular polygon. Draw AE, the longest diagonal, and drop perpendiculars on AE from the other angular points of the polygon. The polygon is thus divided into figures which are either right-angled triangles, rectangles, or trapeziums; and the areas of each of these figures may be readily calculated. |