SECTION IX. OF EQUIVALENCY AND AREA. 311. Equivalent Figures are such as are equal in magnitude. PROPOSITION I. 312. Theorem.-Parallelograms having equal bases and equal altitudes are equivalent. DEM.-Let ABCD and EFGH be two parallelograms having equal bases, BC and FG, and equal altitudes; then are they equivalent. A sides of the one are respectively equal to the three sides of the other. Thus AB = DC, being opposite sides of the same parallelogram. For a like reason, E'B H'C. Also, E'H' = BC AD. From AH' taking E'H', AE' remains, and taking AD, DH' remains. Therefore AE' DH'. These triangles being equal, the quadrilateral ABCH' the triangle AE'B = ABCH' — DH'C. But ABCH' AE'BE'BCH' = EFCH; and ABCH' - DH'C ABCD. Hence, ABCD = Q. E. D. EFCH. = 313. COR.-Any parallelogram is equivalent to a rectangle having the same base and altitude. PROPOSITION II. 314, Theorem.-A triangle is equivalent to one-half of any parallelogram having an equal base and an equal altitude with the triangle. 315. COR. 1.—A triangle is equivalent to one-half of a rectangle having an equal base and an equal altitude with the triangle. 316. COR. 2.-Triangles of equal bases and equal altitudes are equivalent, for they are halves of equivalent parallelograms. PROPOSITION III. 317. Theorem.-The square described on a line is equivalent to four times the square described on half the line, nine times the square described on one-third the line, sixteen times the square on one-fourth the line, etc. DEM.-Let AB be any line. Upon it describe the square ABCD. Bisect AB, as at d, and AD, as at a. Draw de parallel to AD, and ab parallel to AB. Now, the four quadrilaterals thus formed are parallelograms by construction, hence their opposite sides and angles are equal; and as A, B, C, and D are right angles, and Aa = Ad dB = bB etc., the four figures 1, 2, 3, 4, are equal squares. Hence Adoa = ABCD. In like manner it can be shown that the nine figures into which the square on A'B' is divided by draw = C D e ,୯ 7 8 9 4 3 a 6 5 4 2 ing through the points of trisection of the sides, lines parallel to the other sides, are equal squares. Hence A'o', the square on of A'B', is of the square A'B'C'D'. The same process of reasoning can be extended at pleasure, showing that the square on a line is the square of the whole, etc. PROPOSITION IV. 318. Theorem.-A trapezoid is equivalent to two triangles having for their bases the upper and lower bases of the trapezoid, and for their common altitude the altitude of the trapezoid. DEM.-By constructing any trapezoid, and drawing cither diagonal, the student can show the truth of this theorem. PROPOSITION V. 319. Prob.-To reduce any polygon to an equivalent triangle. SOLUTION.-Let ABCDEF be a polygon which it is proposed to reduce to an equivalent triangle. Produce any side, as BC, indefinitely. Draw the diagonal F E EC and DH parallel to it. Draw EH. Now, consider the triangle CDE as cut off from the polygon and replaced by CHE. The magnitude of the polygon will not be changed, since CDE and CHE have the same base CE, and the same altitude, as their vertices lie in DH parallel to EC. From the polygon thus reduced we cut the triangle FHE, and replace it by its equivalent FHI, by drawing the diagonal FH, and the parallel El. In like manner, by drawing FB and the parallel AG, we can replace FBA by its equivalent FCB. Hence, CFI is equivalent to ABCDEF. It is evident that a similar process would reduce a polygon of any number of sides to an equivalent triangle. FIG. 226. AREA. PROPOSITION VI. 320. Theorem.-The area of a rectangle is equal to the product of its base and altitude. DEM.-Let ABCD be a rectangle, then is its area equal to the base AB multiplied by the altitude AC. E If the sides AB and AC are commensurable, take some unit of length, as E, which is contained a whole number of times in each, as five times in AC, and eight times in AB, and apply it to the lines, dividing them respectively into five and eight equal parts. From the several points of division draw lines through the rectangle perpendicular to its sides. The rectangle will be divided into small parallelograms, which are all equal squares, as the angles are all right angles, and the sides all 3 4 5 6 7 8 B equal to each other. Each square is a unit of surface, and the area of the rectangle is expressed by the number of these squares, which is evidently equal to the number in the row on AB, multiplied by the number of such rows, or the number of linear units in AB multiplied by the number in AD. If the two sides of the rectangle are not commensurable, take some very small unit of length which will divide one of the sides, as AC, and divide the rectangle into squares as before; the number of these squares will be the measure of the rectangle, except a small part along one side, not covered by the squares. By taking a still smaller unit, the part left unmeasured by the squares will be still less, and by diminishing the unit of length E, we can make the part unmeasured as small as we choose. It may, therefore, be made infinitely small by regarding the unit of measure as infinitesimal, and consequently is to be neglected.* Hence, in any case, the area of a rectangle is equal to the product of its base into its altitude. Q. E. D. 321. COR. 1.-The area of a square is equal to the second power of one of its sides, as in this case the base and altitude are equal. 322. COR. 2.-The area of any parallelogram is equal to the product of its base into its altitude; for any parallelogram is equivalent to a rectangle of the same base and altitude (313). 323. COR. 3.-The area of a triangle is equal to one-half the product of its base and altitude; for a triangle is one-half of a parallelogram of the same base and altitude (314). 324. COR. 4.-Parallelograms or triangles of equal bases are to each other as their altitudes; of equal altitudes, as their bases; and in general they are to each other as the products of their bases by their altitudes. PROPOSITION VII. 325. Theorem.-The area of a trapezoid is equal to the product of its altitude into one-half the sum of its parallel sides, or, what is the same thing, the product of its altitude and a line joining the middle points of its inclined sides. * This principle may be thus stated: An infinitesimal is a quantity conceived, and to be treated, as less than any assignable quantity; hence, as added to or subtracted from finite quantities, it has no value. Thus, suppose = a, m, n, and a being finite quantities. Let c M represent an infinitesimal; then or or 1 n m±c m m± c a, for to consider it to differ from a by any amount we might name, would be to assign some value to c. + By this is meant the areas of the figures. DEM. In the trapezoid ABCD draw either diagonal, as AC. It is thus D m n B FIG. 228. divided into two triangles, whose areas are together equal to one-half the product of their common altitude (the altitude of the trapezoid), into their bases DC and AB, or this altitude into (AB + DC). Secondly, if ab be drawn bisecting AD and CB, then is ab = (AB + CD). For, through a and b draw the perpendiculars om and pn, meeting DC produced when necessary. Now, the triangles aoD and Aam are equal, since Aa aD, angle o = m, both being right, and angle oaD = Aam being opposite. Whence AmoD. In like manner we may show that Cp nB. Hence, ab = † (op + mn) = (AB + DC); and area ABCD, which equals altitude into (AB + DC), = altitude into ab. Q. E. D. PROPOSITION VIII. 326. Theorem.-The area of a regular polygon is equal to onehalf the product of its apothem into its perimeter. A DEM.-Let ABCDEFG be a regular polygon whose apothem is Oa; then is its area equal to † Oa (AB + BC + CD + DE + EF + FG+ GA). F FIG. 229. E d D Drawing the inscribed circle, the radii Oa, Ob, etc., to the points of tangency, and the radii of the circumscribed circle OA, OB, etc. (264, 265), the polygon is divided into as many equal triangles as it has sides. Now, the apothem (or radius of the inscribed circle) is the common altitude of these triangles, and their bases make up the perimeter of the polygon. Hence, the area = 10a (AB + BC + CD + DE + EF + FG+ GA). Q. E. D. 327. COR.-The area of any polygon in which a circle can be inscribed is equal to one-half the product of the radius of the inscribed circle into the perimeter. The student should draw a figure and observe the fact. It is especially worthy of note in the case of a triangle. See Fig. 60. PROPOSITION IX. 328. Theorem.-The area of a circle is equal to one-half the product of its radius into its circumference. |