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 trianglest 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 ± c represent an infinitesimal; then m mic is to be considered as still equal to n± c' nic' 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 a 7 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 (ABCD). 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 om, 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. g Α 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). G F FIG. 229. E 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 = Oa (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. Circum DEM.-Let Oa be the radius of a circle. scribe any regular polygon. Now the area of this polygon is one-half the product of its apothem and perimeter. Conceive the number of sides of the polygon, indefinitely increased, the polygon still continuing to be circumscribed. The apothem continues to be the radius of the circle, and the perimeter approaches the circumference. When, therefore, the number of sides of the polygon becomes infinite, it is to be considered as coinciding with the circle, and its perimeter with the circumference. Hence the area of the circle is equal to one-half the product of its radius into its circumference. Q. E. D. 329. DEF.-A Sector is a part of a circle included between two radii and their intercepted arc. Similar Sectors are sectors in different circles, which have equal angles at the centre. 330. COR. 1.-The area of a sector is equal to one-half the product of the radius into the arc of the sector. 331. COR. 2.-The area of a sector is to the area of the circle as the arc of the sector is to the circumference, or as the angle of the sector is to four right angles. EXERCISES. 1. What is the area in acres of a triangle whose base is 75 rods and altitude 110 rods? 2. What is the area of a right angled triangle whose sides about the right angle are 126 feet and 72 feet? 3. If 2 lines be drawn from the vertex of a triangle to the base, dividing the base into parts which are to each other as 2, 3, and 5, how is the triangle divided? How does a line drawn from an angle to the middle of the opposite side divide a triangle? 4. Review the exercises on pages 49 and 50, giving the reasons, in each case. SYNOPSIS. Definition. PROP. I. Of parallelograms. Cor. Paral. and rectangle. EXERCISES. SECTION X. OF SIMILARITY. 332. The primary notion of similarity is likeness of form. Two figures are said to be similar which have the same shape, although they may differ in magnitude.* A more scientific definition is as follows: 333. Similar Figures are such as have their angles respectively equal, and their homologous sides proportional. 334. Homologous Sides of similar figures are those which are included between equal angles in the respective figures. * The student should be careful, at the outset, to mark the fact that similarity involves two things, EQUALITY OF ANGLES and PROPORTIONALITY OF SIDES. It will appear that, in the case of triangles, if one of these facts exists, the other does also; but this is not so in other polygons, as is illustrated in PART I. IN SIMILAR TRIANGLES, THE HOMOLOGOUS SIDES ARE THOSE OPPOSITE THE EQUAL ANGLES. PROPOSITION I. 335. Theorem.-Triangles which are mutually equiangular are similar. DEM.-Let ABC and DEF be two mutually equiangular triangles, in which A=D, B=E, and CF; then are the sides opposite these equal angles proportional, and the triangles possess both requisites of similar figures; i. e., they are mutually equiangular and have their homologous sides proportional, and are consequently similar. AA FIG. 231. E To prove that the sides opposite the equal angles are proportional, place the triangle DEF upon ABC, so that F shall coincide with its equal C, CE'=FE, and CD'=FD. Draw AE', and D'B. Since angle CE'D' CBA, D'E' is parallel to AB, and as the triangles D'E'A and D'E'B have a common base D'E' and the same altitude, their vertices lying in a line parallel to their base, they are equivalent (316), Now, the triangles CD'E' and D'E'A, having a common alti-tude, are to each other as their bases (324). Hence, For like reason CD'E' : D'E'A :: CD' : D'A. CD'E' : D'E'B :: CE': E'B. Then, since D'E'A and D'E'B are equivalent, the two proportions have a common ratio, and we may write CD': D'A :: CE' : E'B. By composition CD': CD' + D'A :: CE' : CE' + E'B, or CD': CA :: CE': CB, or FD: CA :: FE : CB. In a similar manner, by applying angle E to B, we can show that 336. COR. 1.-If two triangles have two angles of the one respectively equal to two angles of the other, the third angles being equal (221), the triangles are similar. 337. COR. 2.-A line drawn through a triangle parallel to any side divides the other sides proportionally. Thus D'E' being parallel to AB, it is shown in the proposition that CD': D'A :: CE': E'B. |