To find the number of degrees in the arch ABC 360° : 59° : 44′:: 378.5862: 62.8133 62.8133 length of the arch ABC. 60.25 Ex. 2. What is the area of a fector, when the verfed fine is 5, and chord of half the arch 10? Ex. 3. Required the area of the sector, Anf. 104.720 §. when the chord of half the arch is 10, and the chord of the whole arch 16 feet. Anf. 88.88 feet. Ex. 4. Required the area of a fector of à circle, when the diameter of the circle is 60, and the length of the arch 60 yards. Anf. 900 fq. yards. Ex. 5. Required the area of à sector, when the length of the arch is 156.28 feet, and the diameter 140 feet. Anf. 5469.8 fq. feet. Ex. 6. When the length of the arch is 54, and the radius of the circle 60, required the area. PROBLEM XXIII. Plate 6. fig. 83. To find the area of a fegment of a circle. RULE I. Anf. 1620. Find the area of a sector, whofe arch is the fame with the fegment, by the preceding problem. Then When the chord of half the arch is double the verfed fine, four times the verfed fine is equal to the diameter, and the arch 1200 Then find the area of a triangle, whose two fides are the radii of the sector, and base the chord of the arch. Subtract or add the area of the triangle, according as the fegment is greater or less than a femicircle. RULE 2. Multiply the chord of half the arch by 1, to the product add the chord of the whole arch, multiply this fum by the versed fine, and of the product will be the area of the fegment. EXAMPLE I. What is the area of the fegment ABC, its chord being 60 and radius 50? 360°: 73° 44′ :: 314.16: 64.344 the length of the arch. 64.344 X 25 1608.6 area of the fec. ABCE. DE=20X60=1200 area of tri. ACE 408.6 = area of feg. required. Divide the verfed fine by the diameter, find the quotient in the column of verfed fines, and multiply the correfponding area by the fquare of the diameter for the area of the segment. The example being the fame as before, we have the versed fine equal 10, and diameter 100. In the column of verfed fines find .1 And the correfponding area is Area as before, .040875 10000 408.750000 Ex. 2. Required the area of the fegment, when the arch is 90° and diameter 36 feet. Anf. 92.4696 fq. feet. Ex. 3. What is the area of the fegment of a circle, when the diameter is 25 and verfed fine 9 ? Anf. 159.09. Ex. 4. Required the area of a segment, whofe chord is 32, the radius being 20. Anf. 178.9168. Ex. 5. Required the area of a fegment, its verfed fine being 31, and diameter 50 yards. Anf. 54.1475 Sq. yards. PROBLEM XXIV. To find the area of the cycloid. DEFINITIONS. 1. If the circle ABGE roll on the ftraight line CD, fo that all the points of the circumference be applied to it fucceflively, the point x, that touches the line CD in c, by a motion thus compounded of a circular and rectilineal motion, will defcribe the curve line CBD, which is called the Cycloid. 2. The ftraight line CD is called the bafe. 3. The ftraight line AB, perpendicular to CD, and bifecting it, is called the axis, and is equal to the diameter of the generating circle. 4. The generating circle is that by whofe revolution the curve line is defcribed. 5. The point B is called the vertex. Note. The bafe CD is equal to the circumference of the generating circle, and the cycloid CBD is quadruple of the diameter. Vido Sir Ifaac Newton's Philofophical Discoveries. RULE |