CHAPTER VI. VARIABLE GEOMETRIC MAGNITUDES. GRAPHIC REPRESENTATION. 366. It is often useful to think of a geometric figure as continuously varying in size and shape. = E.g. if a rectangle has a fixed base, say 10 inches long, but an altitude which varies continuously from 3 inches to 5 inches, then the area varies continuously from 3 ∙ 10 30 to 5 ∙ 10 = 50 square inches. We may even think of the altitude as starting at zero inches and increasing continuously, in which case the area starts at zero and increases continuously. From this point of view many theorems may be represented graphically. The graph has the advantage of exhibiting the theorem for all cases at once. For a description of graphic representation see Chapter V of the authors' High School Algebra, Elementary Course. 367. If in the figure AC || BD and OA and OB are commensurable, and if line. OA AC = OB BD then O, C, and D lie in a straight For suppose D not in a line with OC. Pro duce OC and BD to meet at K. Therefore D coincides with K, and O, C and D are in a straight line. 368. THEOREM. The areas of two rectangles having equal bases are in the same ratio as their altitudes. Graphic Representation. For rectangles with commensurable bases and altitudes we have Area = base × altitude. (§ 303) Consider rectangles each with a base equal to b, altitudes h1, h2, h3, etc., each commensurable with b, and areas A, A2, A3, etc. We exhibit graphically the special case where b = 10. Let one horizontal space represent one unit of altitude and one vertical space ten units of area. Thus, the point P2 has the ordinate A = 10 vertical units (representing 100 units of area) and the abscissa h2 = 10 horizontal units. Similarly locate P1 and P ̧ whose abscissas are h1 and hs and ordinates A1 and A2. Using equation (1) and § 367, show that O, P1, P2, P3 ïie in the same straight line. If we suppose that while the base of the rectangle remains fixed, the altitude varies continuously through all values from h2 = 10 to h2 = 20 = 2 × 10, then it must take among other values the value 10 √2. Using 10 √2 as an abscissa, the question is, whether CD is the area ordinate corresponding to it. This area ordinate is not less than CD and commensurable with the base, for in that case the altitude would be less than 10 √2, and for the same reason it is not greater than CD and commensurable with the base. But we can find line-segments less than 10 √2 or greater than 10 √2 and as near to 10 √2 as we please. Hence we conclude that the area ordinate for the rectangle whose altitude is 10 √2 is CD, that is, the point c lies in the line OP¿. In like manner, the point determined by any other abscissa incommensurable with the base is shown to lie on the line OP2. Since the abscissa and ordinate of any point on OP are equal, we have for any altitude, 369. The preceding theorem may also be stated: The area of a rectangle with a fixed base varies directly as its altitude. This means that if A and h are the varying area and altitude respectively, and if A1 and h1 are the area and altitude at any given instant, then 1 The graph representing the relations of two variables when one varies directly as the other is always a straight line. 1. Make a graph to show that the area of two rectangles having equal altitudes are in the same ratio as their bases. 2. Show by a graph that the area of a triangle having a fixed altitude varies as the base, and having a fixed base varies as the altitude. 3. Represent graphically the relation between two line-segments both of which begin at zero, and one of which increases three times as fast as the other. Five times as fast. One half as fast. 4. If areas be represented by the length of a line-segment as in § 368, which question in Ex. 3 applies to the altitude and area of a parallelogram having a fixed base and varying altitude? Which applies to a triangle having a fixed base and varying altitude? 371. THEOREM. The area of a rectangle is equal to the product of its base and altitude. Proof: Using the graphic representations of §§ 368, 370, Ex. 1, we have for all cases, 372. PROBLEM. Make a graphic representation of the theorem: The perimeters of similar polygons are in the same ratio as any two corresponding sides. SOLUTION. First consider the special case of equilateral triangles. On the horizontal axis lay off the lengths of one side of several such triangles, and on the vertical |