gents to the circle through the vertices of the inscribed polygon. If we draw the radii CA, CB, CD, &c. (fig. Fig. 80. 80) to the vertices of the inscribed polygon ABDEFG, and draw through these vertices, perpendicular to radius, the straight lines ab, bil, de, ef, fg, ga, we shall have the regular circumscribed polygon abdefg, of the same number of sides. Remark. The perpendicular CA, CB, &c. drawn from the centre of the regular polygon to one of its sides, (which always bisects that side) is called the apothegm of the polygon. Cm is the apothegm of the inscribed polygon. 148. It is manifest that by increasing the number of sides in the inscribed polygon, we increase its perimeter, which can never be greater than the circumference; and by increasing the number of sides of the circumscribed. pologon, we diminish its perimeter, which can never be less than the circumference. The circumference of the circle being then the limit of this approximation, or the value to which these two magnitudes approach; we say that-The circumference of the circle is equivalent to the perimeter of the regular inscribed (or circumscribed) polygon of an infinite number of sides. Two circles may therefore be considered two similar polygons; whence (139)—The circumferences of two circles are to each other in the ratio of their radii, or of their diameters. Remark. The apothegm of the inscribed or circumscribed polygon of an infinite number of sides must be infinitely near to radius. The consideration of the ratio of the circumference to radius, or to the diameter, we shall defer for the present. PART FIRST. SECTION II.—Of the Measure and Comparison of Plane 149. In discussing the properties of plane figures thus far, we have considered only the lines and the angles, and their several relations to each other, without regard. ing the quantity of surface embraced by the outline or perimeter of the figure. The whole amount of surface or superficies in any geometrical figure is called its area. The word surface, in general, we use to signify superficial extent without regard to quantity. 150. In speaking of figures as equal, we have said that it is an indispensable condition of geometrical equality that the figures compared should coincide by superposition. But it is evident that two figures may have the same amount of surface, and still be very different in form. Such figures we call equivalent. A field of a circular form may have the same superficial extent with another field whose form is quadrangular; we should say that the two fields are equivalent or equal in area. In comparing parallelograms we consider one side as the base of the figure, and the perpendicular distance of this side from the opposite, the height of the figure. Sometimes we call one side the inferior base, and the opposite, the superior base. In triangles one side is taken for the base, and the perpendicular distance of this side from the vertex of the opposite angle, is the height of the triangle. 151. To compare two parallelograms ABCD, ABEF, (fig. 81) whose bases are equal, and heights equal, place Fig. 81 their bases together as in figure 32. On account of their Fig. 82. equal heights and the parallelism of their opposite sides, their superior bases CD, EF, will be in the same straight line CF. We shall have two triangles CAE, DBF, of which the sides CA and DB are equal and parallel, being opposite sides of the same parallelogram (59); AE and BF are equal and parallel for the same reason. The two angles CAE, DBF, having their sides parallel and directed the same way, are equal; and the two triangles CAE, DBF, are equal by the second case of equal trian gles. If from the whole quadrilateral CABF, we take the triangle DBF, there will remain the parallelogram ABDC; and if from the same quadrilateral we take the equal triangle CAE, there will remain the parallelogram ABFE; the two parallelograms must therefore be equivalent. We therefore say-Parallelograms of equal bases and equal heights, are equivalent. 152. As every rectangle is a parallelogram, it follows. that-Every parallelogram is equivalent to a rectangle of an equal base and equal height. 153. We have seen that the diagonal of a parallelogram divides it into two equal triangles (60); the trianFig. 83. gle ABC (fig. 83) is then one half of the parallelogram ABCD; and therefore (152)—Every triangle is equivalent to one half of a rectangle of an equal base and equal height. And consequently-All triangles of equal bases and equal heights, are equivalent. 154. We see that the triangle and parallelogram are both referred to the rectangle for the measure of their arca. If then we can find the exact measure of the rectangle, we have the means of measuring the area of any parallelogram or triangle, and, hence, of any rectilinear figure, as all rectilinear figures may be divided into triangles by drawing diagonals through opposite vertices. 155. To compare two rectangles of the same height and unequal bases, place the less upon the greater so that two equal sides may coincide, and the base CF lie Fig. 94. along the base CD (fig. 84). Fig. 85. If the bases are commensurable, divide them by a common measure; and through the points of division, draw lines parallel to the side CA. Suppose this common measure to be contained eight times in AB, and five times in AE; these lines parallel to CA will divide the larger rectangle into eight equal rectangles, of which the smaller rectangle AECF will contain five. This will give us the ABCD AB ; that is, the two rectangles are proportion AECF AE' to each other in the ratio of their bases. If the bases AB, AE are incommensurable, (fig. 85) divide the larger AB by a measure of the smaller. Suppose it to be contained a certain number of times with the indefinitely small remainder B b, less than any assignable magnitude, which denote by d; and draw through b a straight line parallel to BD. Let us denote the difference between the rectangle whose base is AB and the rectangle whose base is Ab by m; as Ab and AE are commen surable, we shall have the proportion AB AE d ABCD m AECF If we apply the measure of AE one time more than it is contained in AB, it will reach beyond B to b'; and suppose this excess B b', which we denote by d', to be less than any assignable magnitude, (since we can divide AE into parts so small that this excess shall be as little as we please), if we draw through b' a line parallel to BD, meeting CD produced, we shall have a rectangle which exceeds the rectangle ABCD by an indefinitely small excess which we designate by m'. According to what is proved in the preceding part of this article, we have A B C D + m' AB + d' We see, therefore, that in comparing these two rectangles and their bases, if any magnitude however small be added to the rectangle ABCD, a corresponding magnitude must be added to its base AB to preserve the proportion; and if any magnitude however small be subtracted from the rectangle ABCD, a corresponding magnitude must be subtracted from its base to preserve the proportion; it therefore follows necessarily, that if nothing be added to, or subtracted from the rectangle, nothing must be added to, or subtracted from its base, to preserve the proportion; and we ABCD AB consequently have AECF AF; that is, the rectangles are in the ratio of their bases though the bases are incommensurable. We therefore say-Two rectangles of the same height, are to cach other in the ratio of their bases. 156. As the side AC might have been taken as the common base of the two rectangles whose heights are AB and AE; we have from the same proportion ABCD AB this general rule also-Two rectangles AF AECF of equal bases are to each other in the ratio of their heights. 157. As a triangle is equivalent to half the rectangle of the same base and height, it follows from the last two articles, that-Two triangles, of equal heights, are to Fig. 86. each other in the ratio of their bases; and two triangles of equal bases, are to each other in the ratio of their heights. 1158. If we place two rectangles (fig. 86) so as to have the point C a vertex in each, and their sides parallel; and produce BA and FE to meet in H; we shall have another rectangle ACEH, with which each of the others may be compared; these will give the following proportions ABCD CD ACEH AC ACEH CE' CEFG (155) = If we multiply CG ABCD × ACEH these ratios in order, it will give CD X AC will give ACEH X CEFG CEX CG; but in the first ratio of the proportion, the 159. It is usual to estimate areas by square feet, square yards, square rods, &c. By a square foot, we mean a square whose side is one foot; by a square yard, a square whose side is one yard, &c. If we have a rectangular floor whose length is six yards, and breadth five yards, in order to ascertain its area let us divide the end Fig. 87. of the room AC (fig. 87) into yards, and through each division draw lines parallel to AB; this will divide the floor into five equal rectangles, each a yard in width and six yards long; if then we divide the line AB into portions of one yard each, and through these points of division draw lines parallel to A a, we shall divide the rectangle AB ab into six equal squares whose side is one yard; each of the other rectangles abcd, &c. will contain six square yards; so that the whole rectangle will contain five times six square yards, or thirty square yards. This might have been at once obtained by multiplying the number expressing the linear yards in AB, by the number of linear yards in AC. If we would have the area of a rectangle in square |