A circular zone is a figure contained between two parallel chords and the intercepted arcs. To find the area of a circular zone : Find the area of the segment daebc, from which subtract the segment aeb, and the remainder will be the area of the zone. Or, find the area of the trapezoid dabe, to which add twice the area of the segment bc, or ad, and the sum will be the area of the zone. EXAMPLES. 1. What is the area of a circular zone, the chords which bound it on two sides being 20 and 30, and the perpendicular ap, between them, 12, and the chord be 16, and the height of its arc 2? Ans. 334.72. 2. What is the area of a circular zone, the parallel chords of which are 90 and 50, and the distance between them 20? Ans. 1474. To find the diameter of the circle in example second :-Mul tiply the sum of the chords, ab and cd, by their difference, and divide the product by four times the distance between them, or four times ap, and to the quotient add ap, or 20, and the sum is ag; then add the square of ag to the square of the chord ab, and extract the square root of the sum of the squares, and said root will be the diameter, viz. 102.956; and half the difference between ab and dc, viz. 20, will be dp; and the square of radius, minus the square of half the chord ad, will give the square of the apothegm falling on the centre of the chord ad, the root of which will be the apothegm. The chord ad is 28.2825, and the apothegm is 49.4975, which subtracted from radius, leaves the height or versed sine of the are ad, or be 1.9805; and the length of the arc ad 28.64, nearly the area of the segment ad = 37.1649, which, doubled and added to the area of the trapezoid, dabc, gives the area of the zone. = = 3. The chords of a circular zone are 30 and 48, and the distance between them 13; required its area. Ans. 534.188. 1. In a lunar surface the chord ab is 88, the height of the segment aeb is 20, and the height of the segment acb, above the chord ab, is 40; find the area of the lune. Ans. 1452. The radius of the circle, the height of whose arc is 20, I find to be 58.4, and the chord of half the arc of this segment is 48.332, and the length of the arc 99.556, and the area of the segment 1,218, nearly. The radius of the circle, the height of whose segment is 40, I find to be 44.2, and the length of the chord of half the arc 59.464, and the length of the arc 129.237, and its area 2,670, which area, minus the area of the other segment, leaves the area of the crescent. 2. The chord is 48, and the heights, or versed sines, of the two segments are 7 and 18; what is the area? Ans. 405.87. 28. MENSURATION OF THE CONIC SECTIONS. The conic sections are three curves, the parabola, ellipse, and hyperbola. An ellipse is produced by cutting a cone by a plane passing obliquely through its opposite sides. To draw an ellipse :-Draw the major axis, AB, of any required length, and through its centre and at right angles to it, draw the minor axis, CD, of any required length; then take half of the major axis between the dividers, and having placed one foot on the extremity of the minor axis at C, strike an arc cutting the major axis at two points, a and c, and these two points will be the foci of the ellipse. Drive a pin or nail into each of these points, and put a small thread or cord round them, and tie it so that when extended it will just reach the end of the major or minor axis; then with a style or pencil within the thread, extend the string, and describe the ellipse, keeping the thread always stretched, and suffering it to revolve freely about the two pins. The ellipse bears the same ratio to its least circumscribing rectangle, that the circle bears to its least circumscribing square. Therefore, to find the area of an ellipse : Multiply the major by the minor axis, and that product by .7854. To find the circumference of an ellipse : Add together the squares of the major and minor axes, and extract the square root of this sum, and to double this root add one-third of the minor axis. Or, Find the square root of half the sum of the squares of the two diameters, and multiply this root by 3.1416, and the product will be the circumference, nearly; or, the mean between the results found by these two rules, will commonly be a still nearer approximation. An ordinate to either axis, is a line falling perpendicularly upon it from any point in the curve; and this line produced to meet the curve on the other side of the axis, is called a double ordinate; and each of the segments, or parts, into which the ordinate divides the axis, is called an absciss, or abscissa. Thus, eg is an ordinate, and Ag and Bg are the abscisses. The parameter is the same as the double ordinate drawn through the focus; and this line is always a third proportional to the axis on which it falls and the other axis. When the two axes and an absciss are given, to find the ordinate : Say: As the square of the major axis is to that of the minor, so is the product of the two abscisses into each other, to the square of the required ordinate. When the two axes are given and an ordinate, to find the abscisses : Say: As the square of the minor axis is to the square of the major axis, so is the product of the sum and difference of the semi-minor axis and the ordinate, to the distance of the ordinate from the centre of the ellipse; which distance being added to, and subtracted from the semi-major axis, will give the greater and less abscisses. When the minor axis, an ordinate, and an absciss, are given, to find the major axis:— Find the square root of the difference of the squares of the semi-minor axis and the ordinate, and, according as the less or greater absciss is given, add this root to or subtract it from the semi-minor axis; then, as the square of the ordinate is to the product of the absciss and minor axis, so is the sum or difference, found above, to the major axis. When the major axis, an ordinate, and one of the abscisses, are given, to find the minor axis : Find the other absciss; then the product of the two abscisses is to the square of the ordinate, as the square of the major axis to that of the minor axis. When the two axes are given to find the eccentricity of the ellipse, that is, the distance of the foci from the minor axis : - Extract the square root of the difference of the squares of the two axes, and half of said root will be the eccentricity. EXAMPLES. 1. What is the area of an ellipse, whose major or transverse axis is 36 feet, and its minor or conjugate axis is 28 feet? Ans. 791.68 square feet. To solve this and all similar examples by the sliding rule :— Place the major axis found on C over the same on D; then find the minor axis on C, and under it you will find the diameter of a circle on D, whose area is equal to that of the ellipse, viz. 31.74; then place .7854 found on C, over 1 on D, over the diameter of the equal circle, viz. 31.74 found on D, will be found the area on C. and 2. What is the area of an ellipse, whose major axis is 88 rods, and its minor 72? Ans. 31.1 acres, nearly. |