ratio in the case of commensurable segments. It will be sufficient for our purpose to illustrate the process by a special case. Let the student construct carefully a square ABCD, whose side AB measures 10 cm., and draw the diagonal AC. If he measures the diagonal, he will find that its length is between 14 cm. and 15 cm. The ratio of the diagonal to the side is therefore said to 14 15 AC be between and or is between 1.4 and 1.5. Now sup10' AB 10 pose AB to be divided into 100 equal parts (each equal to one millimeter); it will be found that the diagonal is longer than AC 141 mm., but shorter than 142 mm. The ratio is, therefore, AB pose this process could be continued indefinitely. If AB is divided into 1000 equal parts, and these parts applied to AC, beginning at A, the point C must fall on a perfectly definite one of these parts. It would be found that now we should place the AC ratio between 1.414 and 1.415; if AB is divided into 10,000 AB AC equal parts, we should find to lie between 1.4142 and 1.4143, and so on. AB These pairs of numbers differ successively by .1, by .01, by .001, by .0001. This process then makes it possible to define the ratio AB CD to as high a degree of approximation as may be desired, since the process may be thought of as continued indefinitely. Such a process defines a definite number, which is the ratio AC sought. 340. It should be noted that the ratio of two commensurable segments is a rational number, i.e. either an integer or a fraction in which the numerator and denominator are both integers; whereas the ratio of two incommensurable segments is an irrational number, i.e. a number which is not equal to any rational number, but which may be represented approximately by a rational number to as high a degree of approximation as may be desired. Thus the ratio considered above: diagonal of square as will appear later (§ 370). = √2, 341. Measurement of lengths. The process described in § 339 may clearly be applied to any two line segments. The word ratio is then defined for any two line segments. We are thus led to the following definition: The measure of a line segment a in terms of a given segment u chosen as a unit is the ratio of the given segment a to the unit segment u. Quite similar considerations govern D C I 342. Ratio of two areas. the notion of the ratio of two areas and the measurement of areas. We begin by comparing the area of a square with that of a rectangle. K Consider the square AHIK and the rectangle ABCD (Fig. 160). Suppose the side AH of the square and the sides AB, AD of the rectangle have a common measure AL, which is contained exactly m times in AH and n times in AB and p times in AD. (In the figure m = 2, n=4, p=3.) It is then apparent by drawing lines parallel to the sides as shown in the figure that the rectangle can be divided into n⚫p(4.3 squares of side AL and that the square will contain exactly m2 (2.2=4) of these smaller squares. Here the smaller square A H B FIG. 160. = 12) is evidently a common measure of the rectangle and the square and we write the ratio Here, by virtue of the fact that a common measure exists, the rectangle and square are said to be commensurable. It is not necessary at this point to discuss the ratio of incommensurable areas. 343. Measurement of areas. As a unit of measure for measuring areas we use a square. If the unit of length is the inch, the corresponding unit of area is a square whose side measures one inch and is called the square inch; in a similar way we obtain as units the square foot, the square yard, the square mile, etc. To find the measure of an area in terms of a given unit area is to find the ratio of the area to the unit area. Thus in Fig. 160, if AH is 1 in., the area of the rectangle is 12 = 3 square inches. 344. Area of a rectangle. I. Commensurable sides. It is now readily seen that, if the sides of a rectangle are commensurable with the unit of length, the area of the rectangle in terms of the corresponding square unit is equal to the product of the measures of two adjacent sides; i.e. if the sides measure b units and a units, respectively, the area of the rectangle is b. a square units. In the example of § 342, using AH as the unit of length, the measures of the sides AB= are and AD= m 2 p 3 respectively, the segment AL being the common measure of the unit AH and the sides AB and AH. The unit square on AH, contains m2 (=4) of the squares on AL, the rectangle contains np(12) of these squares; the area of the rectangle is, there This proves the statement at the beginning of this section. II. Incommensurable sides. If the sides (one or both) of the rectangle are incommensurable with the unit of length, the situation is not so simple. But it can be shown that in this case also the area of the rectangle is given by the product of the two sides. Consider for example (Fig. 161) a rectangle ABCD whose sides are AD = √5 = 2.2361 and AB = √11 = 3.3166 ... D 2 1 2.3 2.2 From the figure it is clear that the area of the rectangle ABCD lies between the areas of the rectangle whose sides are 2 and 3 and the rectangle whose sides are 3 and 4, i.e. between 6 and 12. If the unit of measure be divided into 10 equal parts, the sides of our rectangle are seen to lie between 2.2 and 2.3 and between 3.3 and 3.4, respectively, and the area sought lies between 2.2 x 3.3 7.2 and 2.3 x 3.47.8. The difference between these two areas is indicated in the figure by the doubly shaded strip. As we take closer and closer rational approximations to the sides of the rectangle we obtain successively pairs of rectangles between the areas of which in every case the area of the rectangle ABCD must lie, and, moreover, the rectangles which form these successive pairs differ by a smaller and smaller amount, as the process proceeds. A LINEAR 3.3 3.1 2 3 B FIG. 161. APPROXIMATE AREA Reference to the adjoining table will show explicitly how this takes place. The first column of this table gives the fractional part of the unit used as an approximate common measure; the second column gives in each case the area of a rectangle less than that of ABCD, the third column, the area of a rectangle greater than that of ABCD. It will be seen that the pairs of corresponding areas in the last two columns agree first in the units' place, then in first decimal place, then in the second decimal place, and so on. 1. 6. 12. 0.1 7.2 7.8 0.01. We conclude, therefore, that the area of the rectangle ABCD is obtained by taking the product of the measures of the two sides, even when one or both of the sides are incommensurable with the unit of length. This is the meaning to be attached to the following theorem : 345. THEOREM. The area of any rectangle is equal to the product of its base by its altitude. 346. Equal polygons. Two polygons which have the same area are said to be equal. Equal polygons are not necessarily congruent. Thus a rectangle whose adjacent sides are 4 and 16 is equal to a square whose side is 8, for each has the area 64 (§ 345); but obviously they are not congruent. 347. Base and altitude of a parallelogram. Since a parallelogram is a special form of trapezoid, any side of the parallelogram may be designated as the base; the (corresponding) altitude is then the perpendicular distance between the base and the opposite side (§§ 228, 229). 348. THEOREM. The area of a parallelogram is equal to the product of its base by its altitude. Given To prove the parallelogram ABCD with the base b and the altitude h. that the area of parallelogram ABCD=bh. Proof. 1. At A and B erect perpendiculars, intersecting CD (produced if necessary) in F and E, respectively. 2. Then ABEF is a rectangle with base b and alti 5. But the area of rectangle ABEF = bħ. Why? Why? Why? Why? § 345 Why? 349. COROLLARY 1. Parallelograms which have equal bases and altitudes are equal (not necessarily congruent). 350. COROLLARY 2. If two parallelograms have equal bases, the ratio of their areas is equal to the ratio of their altitudes; if they have equal altitudes, the ratio of their areas is equal to the ratio of their bases. |