This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1862 Excerpt: ... have a common subtangent on the diameter VS9 because these points, when referred to this diameter and the tangent at its vertex, have the same abscissa, FJV, (Cor. 3, Prop. 6). The point M is therefore common to the two tangents and the diameter VS produced. By this construction we have formed the trapezoid PQQ'P' within, and the triangle TMT without, the parabola, and we will now compare the areas of these figures. From iV'draw NL parallel to PQ, and from Q draw QO perpendicular to P'Q', and let us denote the angle YV'Q that the tangent at Vf makes with the diameter V1 Q by m. By the corollary just referred to we have FFgand V'T'=V'Q'. Whence T T= Q' Q; and because N is the middle point of PP1 we also have 2 Therefore (Th. 34, B. I, Geom.,) the area of the trapezoid PQQP is measured by NLx QO=NLx Q'Q$m.m=Q'QxNLsm.m. ButiVZ/ sin.m is equal to the perpendicular let fall from N upon Qf Q which is equal to that from M upon the same line. Hence the area of the triangle TMT is measured T TxNLzm.m=lQ'QxNLsm.m. The area of the trapezoid is, therefore, twice that of the triangle. If another point be taken between Pf and V% and we make with reference to it and Pf the construction that has just been made with reference to P' and P, we shall have another trapezoid within, and triangle without, the parabola, and the area of the trapezoid will be twice that of the triangle. Let us suppose this process continued until we have inscribed a polygon in the parabola between the limits P and V7; then, if the distance of the consecutive points P, P', etc., be supposed indefinitely small, it is evident that the sum of the trapezoids will become the interior curvilinear area PP'V'Q, and the sum of the triangles the exterior curvilinear area TPV V. Since any one of these trapezoi...

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